Submitting multiple database updates via PHP & JQuery

Question

I'm stuck with trying to process multiple mySQL updates at the same time. I have 4 select/optiion boxes that pull entries from a db table. I want to be able to update the db onChange using JQuery. I have managed to get this working with one select module but as soon as I add more it spins out. I know that the main bad code is in db_submit.php but really not sure how else to write it. I know there has to be a cleaner way to do this.

FORM PAGE- INPUT.PHP

<html>
<head>
<script src="../assets/scripts/jquery-2.0.3.min.js"></script>
<script> 
    function updateDb() {
     $.post("db_submit.php", $("#console").serialize());
    }
</script>
<?php
include 'db_connect.php';
?>
</head>

<body>
<form id="console">
    <select id="frame1" name="frame1" onChange="updateDb()">
        <option value="">Select Channel</option>
            <?php
            $result = mysqli_query($con,"SELECT * FROM feedContent");
            while ($row = mysqli_fetch_array($result)) {
                echo "<option value='" . $row['url'] . "'>" . $row['name'] . "</option>";
            }
            ?>
    </select>
    <select id="frame2" name="frame2" onChange="updateDb()">
        <option value="">Select Channel</option>
            <?php
            $result = mysqli_query($con,"SELECT * FROM feedContent");
            while ($row = mysqli_fetch_array($result)) {
                echo "<option value='" . $row['url'] . "'>" . $row['name'] . "</option>";
            }
            ?>
    </select>
    <select id="frame3" name="frame3" onChange="updateDb()">
        <option value="">Select Channel</option>
            <?php
            $result = mysqli_query($con,"SELECT * FROM feedContent");
            while ($row = mysqli_fetch_array($result)) {
                echo "<option value='" . $row['url'] . "'>" . $row['name'] . "</option>";
            }
            ?>
    </select>
    <select id="frame4" name="frame4" onChange="updateDb()">
        <option value="">Select Channel</option>
            <?php
            $result = mysqli_query($con,"SELECT * FROM feedContent");
            while ($row = mysqli_fetch_array($result)) {
                echo "<option value='" . $row['url'] . "'>" . $row['name'] . "</option>";
            }
            ?>
    </select>
</form>
</body>
<?php
mysqli_close($con);
?>
</html>

PROCESSING PAGE- DB_SUBMIT.PHP

<?php
include 'db_connect.php';   
$frame1= mysqli_escape_String($con,$_POST['frame1']);
$frame2= mysqli_escape_String($con,$_POST['frame2']);
$frame3= mysqli_escape_String($con,$_POST['frame3']);
$frame4= mysqli_escape_String($con,$_POST['frame4']);

$query = "UPDATE frameContent SET url='".$frame1."' WHERE name='frame1'";
$query = "UPDATE frameContent SET url='".$frame2."' WHERE name='frame2'";
$query = "UPDATE frameContent SET url='".$frame3."' WHERE name='frame3'";
$query = "UPDATE frameContent SET url='".$frame4."' WHERE name='frame4'";
mysqli_query($con,$query);

mysqli_close($con);
?>

I know that constantly setting the $query variable is causing problems but I'm not sure how else I can do this in the one page. Any help would be much appreciated.

Thanks!

vee · Accepted Answer · 2013-12-19 06:01:14Z

3

First of all make sure the $queries are concatenated, then terminate each query with a semi-colon. After these you can use mysqli_multi_query to execute all four updates in one call from php.

$query = "UPDATE frameContent SET url='".$frame1."' WHERE name='frame1';";
$query .= "UPDATE frameContent SET url='".$frame2."' WHERE name='frame2';";
$query .= "UPDATE frameContent SET url='".$frame3."' WHERE name='frame3';";
$query .= "UPDATE frameContent SET url='".$frame4."' WHERE name='frame4';";
mysqli_multi_query($con,$query);

answered Dec 19, 2013 at 6:01

vee

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2 Comments

adroxx Over a year ago

This worked in Chrome but Safari seems to crash it. When submitting in Safari it clears all the rows in the db. Any thoughts?

adroxx Over a year ago

Please ignore my last comment. I found the problem. I had a typo in my code. Thanks again! :)

Aljie · Accepted Answer · 2013-12-19 07:04:45Z

I think this might help :) but there's just a little changes within your codes:

<html>
<head>
<script src = "js/jquery-1.10.1.js"></script>
<script> 
function updateDb() 
{
 // this var id will store all your 4 combobox values in an array
  var id = [{val1: $("#frame1").val()},
           {val1: $("#frame2").val()},
           {val1: $("#frame3").val()},
           {val1: $("#frame4").val()}];

 //this .post will submit all data to db_submit.php
   $.post("db_submit.php",{id:id}, function(data)
  {
           alert(data);
  });
</script>
<?php
  include 'db_connect.php';
?>
</head>
<body>
<select id="frame1" name="frame1">
    <option value="">Select Channel</option>
        <?php
        $result = mysqli_query($con,"SELECT * FROM feedContent");
        while ($row = mysqli_fetch_array($result)) {
            echo "<option value='" . $row['url'] . "'>" . $row['name'] . "</option>";
        }
        ?>
</select>
<select id="frame2" name="frame2">
    <option value="">Select Channel</option>
        <?php
        $result = mysqli_query($con,"SELECT * FROM feedContent");
        while ($row = mysqli_fetch_array($result)) {
            echo "<option value='" . $row['url'] . "'>" . $row['name'] . "</option>";
        }
        ?>
</select>
<select id="frame3" name="frame3">
    <option value="">Select Channel</option>
        <?php
        $result = mysqli_query($con,"SELECT * FROM feedContent");
        while ($row = mysqli_fetch_array($result)) {
            echo "<option value='" . $row['url'] . "'>" . $row['name'] . "</option>";
        }
        ?>
</select>
<select id="frame4" name="frame4">
    <option value="">Select Channel</option>
        <?php
        $result = mysqli_query($con,"SELECT * FROM feedContent");
        while ($row = mysqli_fetch_array($result)) {
            echo "<option value='" . $row['url'] . "'>" . $row['name'] . "</option>";
        }
        ?>
</select>

<input type="button" value="Submit" onClick="updateDb()"/>
</body>
<?php
   mysqli_close($con);
?>
</html>

And in your DB_SUBMIT.PHP

 <?php
 include 'db_connect.php';   
 $frame1= mysqli_escape_String($_POST['id'][0]['val1']);
 $frame2= mysqli_escape_String($_POST['id'][1]['val1']);
 $frame3= mysqli_escape_String($_POST['id'][2]['val1']);
 $frame4= mysqli_escape_String($_POST['id'][3]['val1']);

 $query = mysqli_query("UPDATE frameContent SET url='$frame1' WHERE name='frame1'");
 $query = mysqli_query("UPDATE frameContent SET url='$frame2' WHERE name='frame2'");
 $query = mysqli_query("UPDATE frameContent SET url='$frame3' WHERE name='frame3'");
 $query = mysqli_query("UPDATE frameContent SET url='$frame4' WHERE name='frame4'");

 echo "Data was Successfully updated";
 mysqli_close($con);
 ?>

I just add a button there for convenience, but if you dont want it just delete it and put back the onChange on every combocboxes that you have :)

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