Inserting Data in multiple tables In database Using If statement and array
i'm beginner in PHP and i having problems with my code..i want to make a little fee and income system.its a web based app.. i want to insert data in two different tables 1) fees and 2)income using options..here im taking two steps (1) select data from database (2) inserting in data base.
1st step is working good problem with 2nd step the data is inserted in both 1) fees and 2)income, not all records saved i click for saving 3 records but there is only one or two record save.
select data from database
<?php
if(isset($_POST['classes'])){
$cla=$_POST['classes'];}$sql = "SELECT class,std_name,roll_no,monthly_fee from students where class = '$cla'";
$result = $conn->query($sql);if ($result->num_rows > 0) {$counter = 0;while($row = $result->fetch_assoc()) {?><tr><td><?php echo $row["roll_no"]; ?><input type="hidden" value="<?php echo $row["roll_no"]; ?>" name="roll_no"></td><td><?php echo $row["std_name"]; ?><input type="hidden" value="<?php echo$row["std_name"]; ?>" name="std_name"></td><td><?php echo $row["class"]; ?><input type="hidden" value="<?php echo $row["class"]; ?>" name="class"></td><td><?php echo $row["monthly_fee"]; ?><input type="hidden" value="<?php echo$row["monthly_fee"]; ?>" name="monthly_fee"></td><td><select name="status[<?php echo $counter; ?>]"><option >--Select Status--</option>
<option value="Paid">Paid</option>
<option value="Unpaid">Unpaid</option>
Insert into Database
i think the problems is with this code
here i want to store data with condition
if (isset($_POST['submit'])) {
foreach ($_POST['status'] as $state => $status) {
$roll_no = $_POST['roll_no'][$state];
$std_name = $_POST['std_name'][$state];
$class = $_POST['class'][$state];
$monthly_fee = $_POST['monthly_fee'][$state];
$stat = $_POST['status'][$state];
$date = date('Y-m-d H:i:s');
$sql = "INSERT INTO fees(std_id,std_name,std_class,fee_status,monthly_fee,taken_date) VALUES( '$roll_no','$std_name','$class','$stat','$monthly_fee','$date')";
if ($state == 'Paid') {
$sql = "INSERT INTO income(std_fee) values ('$monthly_fee')";if ($conn->query($sql) === TRUE) {$flag = 1;}else{echo "error".$sql."<br>".$conn->error;} }}}?>
i hope you will understand
php sql arrays html5
asked Nov 19 '18 at 19:44
Atif RajaAtif Raja
126
add a comment |
i'm beginner in PHP and i having problems with my code..i want to make a little fee and income system.its a web based app.. i want to insert data in two different tables 1) fees and 2)income using options..here im taking two steps (1) select data from database (2) inserting in data base.
1st step is working good problem with 2nd step the data is inserted in both 1) fees and 2)income, not all records saved i click for saving 3 records but there is only one or two record save.
select data from database
<?php
if(isset($_POST['classes'])){
$cla=$_POST['classes'];}$sql = "SELECT class,std_name,roll_no,monthly_fee from students where class = '$cla'";
$result = $conn->query($sql);if ($result->num_rows > 0) {$counter = 0;while($row = $result->fetch_assoc()) {?><tr><td><?php echo $row["roll_no"]; ?><input type="hidden" value="<?php echo $row["roll_no"]; ?>" name="roll_no"></td><td><?php echo $row["std_name"]; ?><input type="hidden" value="<?php echo$row["std_name"]; ?>" name="std_name"></td><td><?php echo $row["class"]; ?><input type="hidden" value="<?php echo $row["class"]; ?>" name="class"></td><td><?php echo $row["monthly_fee"]; ?><input type="hidden" value="<?php echo$row["monthly_fee"]; ?>" name="monthly_fee"></td><td><select name="status[<?php echo $counter; ?>]"><option >--Select Status--</option>
<option value="Paid">Paid</option>
<option value="Unpaid">Unpaid</option>
Insert into Database
i think the problems is with this code
here i want to store data with condition
if (isset($_POST['submit'])) {
foreach ($_POST['status'] as $state => $status) {
$roll_no = $_POST['roll_no'][$state];
$std_name = $_POST['std_name'][$state];
$class = $_POST['class'][$state];
$monthly_fee = $_POST['monthly_fee'][$state];
$stat = $_POST['status'][$state];
$date = date('Y-m-d H:i:s');
$sql = "INSERT INTO fees(std_id,std_name,std_class,fee_status,monthly_fee,taken_date) VALUES( '$roll_no','$std_name','$class','$stat','$monthly_fee','$date')";
if ($state == 'Paid') {
$sql = "INSERT INTO income(std_fee) values ('$monthly_fee')";if ($conn->query($sql) === TRUE) {$flag = 1;}else{echo "error".$sql."<br>".$conn->error;} }}}?>
i hope you will understand
php sql arrays html5
asked Nov 19 '18 at 19:44
Atif RajaAtif Raja
126
2
Can you format your code properly as it is currently loads of commands on single lines.
– Nigel Ren
Nov 19 '18 at 19:57
add a comment |
i'm beginner in PHP and i having problems with my code..i want to make a little fee and income system.its a web based app.. i want to insert data in two different tables 1) fees and 2)income using options..here im taking two steps (1) select data from database (2) inserting in data base.
1st step is working good problem with 2nd step the data is inserted in both 1) fees and 2)income, not all records saved i click for saving 3 records but there is only one or two record save.
select data from database
<?php
if(isset($_POST['classes'])){
$cla=$_POST['classes'];}$sql = "SELECT class,std_name,roll_no,monthly_fee from students where class = '$cla'";
$result = $conn->query($sql);if ($result->num_rows > 0) {$counter = 0;while($row = $result->fetch_assoc()) {?><tr><td><?php echo $row["roll_no"]; ?><input type="hidden" value="<?php echo $row["roll_no"]; ?>" name="roll_no"></td><td><?php echo $row["std_name"]; ?><input type="hidden" value="<?php echo$row["std_name"]; ?>" name="std_name"></td><td><?php echo $row["class"]; ?><input type="hidden" value="<?php echo $row["class"]; ?>" name="class"></td><td><?php echo $row["monthly_fee"]; ?><input type="hidden" value="<?php echo$row["monthly_fee"]; ?>" name="monthly_fee"></td><td><select name="status[<?php echo $counter; ?>]"><option >--Select Status--</option>
<option value="Paid">Paid</option>
<option value="Unpaid">Unpaid</option>
Insert into Database
i think the problems is with this code
here i want to store data with condition
if (isset($_POST['submit'])) {
foreach ($_POST['status'] as $state => $status) {
$roll_no = $_POST['roll_no'][$state];
$std_name = $_POST['std_name'][$state];
$class = $_POST['class'][$state];
$monthly_fee = $_POST['monthly_fee'][$state];
$stat = $_POST['status'][$state];
$date = date('Y-m-d H:i:s');
$sql = "INSERT INTO fees(std_id,std_name,std_class,fee_status,monthly_fee,taken_date) VALUES( '$roll_no','$std_name','$class','$stat','$monthly_fee','$date')";
if ($state == 'Paid') {
$sql = "INSERT INTO income(std_fee) values ('$monthly_fee')";if ($conn->query($sql) === TRUE) {$flag = 1;}else{echo "error".$sql."<br>".$conn->error;} }}}?>
i hope you will understand
php sql arrays html5
asked Nov 19 '18 at 19:44
Atif RajaAtif Raja
126
i'm beginner in PHP and i having problems with my code..i want to make a little fee and income system.its a web based app.. i want to insert data in two different tables 1) fees and 2)income using options..here im taking two steps (1) select data from database (2) inserting in data base.
1st step is working good problem with 2nd step the data is inserted in both 1) fees and 2)income, not all records saved i click for saving 3 records but there is only one or two record save.
select data from database
<?php
if(isset($_POST['classes'])){
$cla=$_POST['classes'];}$sql = "SELECT class,std_name,roll_no,monthly_fee from students where class = '$cla'";
$result = $conn->query($sql);if ($result->num_rows > 0) {$counter = 0;while($row = $result->fetch_assoc()) {?><tr><td><?php echo $row["roll_no"]; ?><input type="hidden" value="<?php echo $row["roll_no"]; ?>" name="roll_no"></td><td><?php echo $row["std_name"]; ?><input type="hidden" value="<?php echo$row["std_name"]; ?>" name="std_name"></td><td><?php echo $row["class"]; ?><input type="hidden" value="<?php echo $row["class"]; ?>" name="class"></td><td><?php echo $row["monthly_fee"]; ?><input type="hidden" value="<?php echo$row["monthly_fee"]; ?>" name="monthly_fee"></td><td><select name="status[<?php echo $counter; ?>]"><option >--Select Status--</option>
<option value="Paid">Paid</option>
<option value="Unpaid">Unpaid</option>
Insert into Database
i think the problems is with this code
here i want to store data with condition
if (isset($_POST['submit'])) {
foreach ($_POST['status'] as $state => $status) {
$roll_no = $_POST['roll_no'][$state];
$std_name = $_POST['std_name'][$state];
$class = $_POST['class'][$state];
$monthly_fee = $_POST['monthly_fee'][$state];
$stat = $_POST['status'][$state];
$date = date('Y-m-d H:i:s');
$sql = "INSERT INTO fees(std_id,std_name,std_class,fee_status,monthly_fee,taken_date) VALUES( '$roll_no','$std_name','$class','$stat','$monthly_fee','$date')";
if ($state == 'Paid') {
$sql = "INSERT INTO income(std_fee) values ('$monthly_fee')";if ($conn->query($sql) === TRUE) {$flag = 1;}else{echo "error".$sql."<br>".$conn->error;} }}}?>
i hope you will understand
php sql arrays html5
php sql arrays html5
asked Nov 19 '18 at 19:44
Atif RajaAtif Raja
126
asked Nov 19 '18 at 19:44
Atif RajaAtif Raja
126
asked Nov 19 '18 at 19:44
Atif RajaAtif Raja
126
asked Nov 19 '18 at 19:44
Atif RajaAtif Raja
126
asked Nov 19 '18 at 19:44
Atif RajaAtif Raja
126
126
2
Can you format your code properly as it is currently loads of commands on single lines.
– Nigel Ren
Nov 19 '18 at 19:57
add a comment |
2
Can you format your code properly as it is currently loads of commands on single lines.
– Nigel Ren
Nov 19 '18 at 19:57
2
2
Can you format your code properly as it is currently loads of commands on single lines.
– Nigel Ren
Nov 19 '18 at 19:57
Can you format your code properly as it is currently loads of commands on single lines.
– Nigel Ren
Nov 19 '18 at 19:57
add a comment |
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2
Can you format your code properly as it is currently loads of commands on single lines.
– Nigel Ren
Nov 19 '18 at 19:57