python3 flask: how to display the json result in html





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Now I want to send the json data to html, I just display the dictionary data(json), How to display the value of dictionary in the html.
The python code is:



@app.route('/upload', methods=['GET', 'POST'])
def upload():
if request.method == 'POST':
img_file = request.files['img_file']
if img_file and allowed_file(img_file.filename):
filename = secure_filename(img_file.filename)
img_file.save(os.path.join(app.config['UPLOAD_FOLDER'], filename))
raw_img_url = './static/asset/uploads/' + filename
result_img_url = './static/asset/output/Line/' + filename

ratio = check_banner(filename)
result_dict = {'result': ratio}

return render_template("upload.html", result_img_url=result_img_url, result=json.dumps(result_dict))
else:
return ''' ok</p> '''
else:
return redirect(url_for('upload'))


The html code:



{% extends "base.html" %}
{% block content %}

<form method="post" action="/upload" enctype="multipart/form-data">
<input type="file" id="img_file" name="img_file" class="col-sm-4">
<input type="submit" value="check" class="btn">
</form>

<p>
{% if result_img_url %}
<img src="{{ result_img_url }}" width='400' height='350'>
{% endif %}
</p>

<ul>

<li>{{result}}</li>

</ul>


{% endblock %}


But the result html is, but I just want to show the value of result :



result html










share|improve this question





























    0















    Now I want to send the json data to html, I just display the dictionary data(json), How to display the value of dictionary in the html.
    The python code is:



    @app.route('/upload', methods=['GET', 'POST'])
    def upload():
    if request.method == 'POST':
    img_file = request.files['img_file']
    if img_file and allowed_file(img_file.filename):
    filename = secure_filename(img_file.filename)
    img_file.save(os.path.join(app.config['UPLOAD_FOLDER'], filename))
    raw_img_url = './static/asset/uploads/' + filename
    result_img_url = './static/asset/output/Line/' + filename

    ratio = check_banner(filename)
    result_dict = {'result': ratio}

    return render_template("upload.html", result_img_url=result_img_url, result=json.dumps(result_dict))
    else:
    return ''' ok</p> '''
    else:
    return redirect(url_for('upload'))


    The html code:



    {% extends "base.html" %}
    {% block content %}

    <form method="post" action="/upload" enctype="multipart/form-data">
    <input type="file" id="img_file" name="img_file" class="col-sm-4">
    <input type="submit" value="check" class="btn">
    </form>

    <p>
    {% if result_img_url %}
    <img src="{{ result_img_url }}" width='400' height='350'>
    {% endif %}
    </p>

    <ul>

    <li>{{result}}</li>

    </ul>


    {% endblock %}


    But the result html is, but I just want to show the value of result :



    result html










    share|improve this question

























      0












      0








      0








      Now I want to send the json data to html, I just display the dictionary data(json), How to display the value of dictionary in the html.
      The python code is:



      @app.route('/upload', methods=['GET', 'POST'])
      def upload():
      if request.method == 'POST':
      img_file = request.files['img_file']
      if img_file and allowed_file(img_file.filename):
      filename = secure_filename(img_file.filename)
      img_file.save(os.path.join(app.config['UPLOAD_FOLDER'], filename))
      raw_img_url = './static/asset/uploads/' + filename
      result_img_url = './static/asset/output/Line/' + filename

      ratio = check_banner(filename)
      result_dict = {'result': ratio}

      return render_template("upload.html", result_img_url=result_img_url, result=json.dumps(result_dict))
      else:
      return ''' ok</p> '''
      else:
      return redirect(url_for('upload'))


      The html code:



      {% extends "base.html" %}
      {% block content %}

      <form method="post" action="/upload" enctype="multipart/form-data">
      <input type="file" id="img_file" name="img_file" class="col-sm-4">
      <input type="submit" value="check" class="btn">
      </form>

      <p>
      {% if result_img_url %}
      <img src="{{ result_img_url }}" width='400' height='350'>
      {% endif %}
      </p>

      <ul>

      <li>{{result}}</li>

      </ul>


      {% endblock %}


      But the result html is, but I just want to show the value of result :



      result html










      share|improve this question














      Now I want to send the json data to html, I just display the dictionary data(json), How to display the value of dictionary in the html.
      The python code is:



      @app.route('/upload', methods=['GET', 'POST'])
      def upload():
      if request.method == 'POST':
      img_file = request.files['img_file']
      if img_file and allowed_file(img_file.filename):
      filename = secure_filename(img_file.filename)
      img_file.save(os.path.join(app.config['UPLOAD_FOLDER'], filename))
      raw_img_url = './static/asset/uploads/' + filename
      result_img_url = './static/asset/output/Line/' + filename

      ratio = check_banner(filename)
      result_dict = {'result': ratio}

      return render_template("upload.html", result_img_url=result_img_url, result=json.dumps(result_dict))
      else:
      return ''' ok</p> '''
      else:
      return redirect(url_for('upload'))


      The html code:



      {% extends "base.html" %}
      {% block content %}

      <form method="post" action="/upload" enctype="multipart/form-data">
      <input type="file" id="img_file" name="img_file" class="col-sm-4">
      <input type="submit" value="check" class="btn">
      </form>

      <p>
      {% if result_img_url %}
      <img src="{{ result_img_url }}" width='400' height='350'>
      {% endif %}
      </p>

      <ul>

      <li>{{result}}</li>

      </ul>


      {% endblock %}


      But the result html is, but I just want to show the value of result :



      result html







      python-3.x web flask






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked Nov 23 '18 at 1:48









      tktktk0711tktktk0711

      611834




      611834
























          1 Answer
          1






          active

          oldest

          votes


















          1














          If you want to pass only one parameter in result, you may change result_dict = {'result': ratio} to result = ratio.



          It will show only the result in the page:



          show only value






          share|improve this answer
























          • OK thanks!, after I do this, I want to do:{% if result <= 0.2 and (size is 1) %} <th> Pass </th> <th> {{ type(result) }} </th> {% elif result = 0.24 and (size is 1) %} <th> Need Check </th> {% (elif result > 0.24 and size is 1) or (size is 0) %} <th> no pass </th> </p>

            – tktktk0711
            Nov 23 '18 at 4:27













          • but there is error: jinja2.exceptions.TemplateSyntaxError: expected token 'name', got 'integer'

            – tktktk0711
            Nov 23 '18 at 4:28











          • @tktktk0711, if this answer solves the queries in scope of the question please accept it. For further queries, add a separate question with details of the template and error traceback. Asking questions in comments and answering them in comments are not advised in Stackoverflow. :)

            – arsho
            Nov 23 '18 at 6:16














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          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1














          If you want to pass only one parameter in result, you may change result_dict = {'result': ratio} to result = ratio.



          It will show only the result in the page:



          show only value






          share|improve this answer
























          • OK thanks!, after I do this, I want to do:{% if result <= 0.2 and (size is 1) %} <th> Pass </th> <th> {{ type(result) }} </th> {% elif result = 0.24 and (size is 1) %} <th> Need Check </th> {% (elif result > 0.24 and size is 1) or (size is 0) %} <th> no pass </th> </p>

            – tktktk0711
            Nov 23 '18 at 4:27













          • but there is error: jinja2.exceptions.TemplateSyntaxError: expected token 'name', got 'integer'

            – tktktk0711
            Nov 23 '18 at 4:28











          • @tktktk0711, if this answer solves the queries in scope of the question please accept it. For further queries, add a separate question with details of the template and error traceback. Asking questions in comments and answering them in comments are not advised in Stackoverflow. :)

            – arsho
            Nov 23 '18 at 6:16


















          1














          If you want to pass only one parameter in result, you may change result_dict = {'result': ratio} to result = ratio.



          It will show only the result in the page:



          show only value






          share|improve this answer
























          • OK thanks!, after I do this, I want to do:{% if result <= 0.2 and (size is 1) %} <th> Pass </th> <th> {{ type(result) }} </th> {% elif result = 0.24 and (size is 1) %} <th> Need Check </th> {% (elif result > 0.24 and size is 1) or (size is 0) %} <th> no pass </th> </p>

            – tktktk0711
            Nov 23 '18 at 4:27













          • but there is error: jinja2.exceptions.TemplateSyntaxError: expected token 'name', got 'integer'

            – tktktk0711
            Nov 23 '18 at 4:28











          • @tktktk0711, if this answer solves the queries in scope of the question please accept it. For further queries, add a separate question with details of the template and error traceback. Asking questions in comments and answering them in comments are not advised in Stackoverflow. :)

            – arsho
            Nov 23 '18 at 6:16
















          1












          1








          1







          If you want to pass only one parameter in result, you may change result_dict = {'result': ratio} to result = ratio.



          It will show only the result in the page:



          show only value






          share|improve this answer













          If you want to pass only one parameter in result, you may change result_dict = {'result': ratio} to result = ratio.



          It will show only the result in the page:



          show only value







          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered Nov 23 '18 at 3:48









          arshoarsho

          4,14411838




          4,14411838













          • OK thanks!, after I do this, I want to do:{% if result <= 0.2 and (size is 1) %} <th> Pass </th> <th> {{ type(result) }} </th> {% elif result = 0.24 and (size is 1) %} <th> Need Check </th> {% (elif result > 0.24 and size is 1) or (size is 0) %} <th> no pass </th> </p>

            – tktktk0711
            Nov 23 '18 at 4:27













          • but there is error: jinja2.exceptions.TemplateSyntaxError: expected token 'name', got 'integer'

            – tktktk0711
            Nov 23 '18 at 4:28











          • @tktktk0711, if this answer solves the queries in scope of the question please accept it. For further queries, add a separate question with details of the template and error traceback. Asking questions in comments and answering them in comments are not advised in Stackoverflow. :)

            – arsho
            Nov 23 '18 at 6:16





















          • OK thanks!, after I do this, I want to do:{% if result <= 0.2 and (size is 1) %} <th> Pass </th> <th> {{ type(result) }} </th> {% elif result = 0.24 and (size is 1) %} <th> Need Check </th> {% (elif result > 0.24 and size is 1) or (size is 0) %} <th> no pass </th> </p>

            – tktktk0711
            Nov 23 '18 at 4:27













          • but there is error: jinja2.exceptions.TemplateSyntaxError: expected token 'name', got 'integer'

            – tktktk0711
            Nov 23 '18 at 4:28











          • @tktktk0711, if this answer solves the queries in scope of the question please accept it. For further queries, add a separate question with details of the template and error traceback. Asking questions in comments and answering them in comments are not advised in Stackoverflow. :)

            – arsho
            Nov 23 '18 at 6:16



















          OK thanks!, after I do this, I want to do:{% if result <= 0.2 and (size is 1) %} <th> Pass </th> <th> {{ type(result) }} </th> {% elif result = 0.24 and (size is 1) %} <th> Need Check </th> {% (elif result > 0.24 and size is 1) or (size is 0) %} <th> no pass </th> </p>

          – tktktk0711
          Nov 23 '18 at 4:27







          OK thanks!, after I do this, I want to do:{% if result <= 0.2 and (size is 1) %} <th> Pass </th> <th> {{ type(result) }} </th> {% elif result = 0.24 and (size is 1) %} <th> Need Check </th> {% (elif result > 0.24 and size is 1) or (size is 0) %} <th> no pass </th> </p>

          – tktktk0711
          Nov 23 '18 at 4:27















          but there is error: jinja2.exceptions.TemplateSyntaxError: expected token 'name', got 'integer'

          – tktktk0711
          Nov 23 '18 at 4:28





          but there is error: jinja2.exceptions.TemplateSyntaxError: expected token 'name', got 'integer'

          – tktktk0711
          Nov 23 '18 at 4:28













          @tktktk0711, if this answer solves the queries in scope of the question please accept it. For further queries, add a separate question with details of the template and error traceback. Asking questions in comments and answering them in comments are not advised in Stackoverflow. :)

          – arsho
          Nov 23 '18 at 6:16







          @tktktk0711, if this answer solves the queries in scope of the question please accept it. For further queries, add a separate question with details of the template and error traceback. Asking questions in comments and answering them in comments are not advised in Stackoverflow. :)

          – arsho
          Nov 23 '18 at 6:16






















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