Populating javascript 2D array from php-mysql

Question

I am new to Javascript/PHP, so have a few basic questions. I am trying to create a javascript array from php. All goes okay till I am reference collumn '0' or 'pid' from my sql table, but as soon as I try to access the next collumn 'Function' the javascript stops working. I have created a testcase with alert (same to be replaced by an array) Would really appreciate all your help.

<?php
 $baseName . "jobs";
 $tableName     =   $baseName . "dbTable";
 # datbase info .done. 
 $con = mysql_connect("localhost",$userName,$password);
 if (!$con){
     echo('Could not connect: ' . mysql_error());
 }
 mysql_select_db($dbName, $con);
 $result = mysql_query("SELECT * FROM " . $tableName);
 $count = 0;
 echo "<table border='1'>
 <tr>
 <th>ID</th>
 <th>Function</th>
 <th>Position</th>
 <th>Location</th>
 <th>Experience</th>
 </tr>";
 while($row = mysql_fetch_array($result))
 {  
 echo "<tr>";
 echo "<td>" . $row['pid'] . "</td>";
 echo "<td>" . $row['Function'] . "</td>";
 echo "<td>" . $row['Position'] . "</td>";
 echo "<td>" . $row['City'] . ', ' . $row['State'] . "</td>";
 echo "<td>" . $row['Experience'] . "</td>";
 echo "</tr>";
?>
<script type="text/javascript">
 alert(<?php echo $row['pid']?>); <!-- this works -->
 **alert(<?php echo $row['Function']?>); <!-- this does not works -->**
</script>
<?php           
 $count++;
}
echo "</table>";
mysql_close($con);
?>

vusan · Accepted Answer · 2012-09-26 11:18:56Z

1

you should have put double quote or single quote on your after alert. like

alert("<?php echo $row['Function']?>");

You get value of first because it is just number Now if it is text you won't get after.

answered Sep 26, 2012 at 11:18

vusan

5,3394 gold badges49 silver badges83 bronze badges

Thanks for your speedy replies, the double quotes is what I needed. I have learned a lot from this forum!

RB1987
– RB1987

09/26/2012 11:42:13
Commented Sep 26, 2012 at 11:42

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MatthewMcGovern · Accepted Answer · 2012-09-26 11:16:04Z

0

What is your SQL table structure? Doing $row['name'] will try and get the data under the column 'name'. It must match exactly for it to be able to return data.

Also, the mysql commands are getting deprecated so look into mysqli or PDOs.

answered Sep 26, 2012 at 11:16

MatthewMcGovern

3,5062 gold badges23 silver badges19 bronze badges

The table headers are: pid|Function|Position|City|Experience, the 'name' should be fine, as the echo's from php code(inside while) work, also if the name is misspelled or incorrect the alert wreturns an empty string, but atleast works. In my case the browser does not throw an alert if I refrence the collumn headers(other than pid)

RB1987
– RB1987

09/26/2012 11:18:48
Commented Sep 26, 2012 at 11:18
Ah I see, then Vusan's answer is most likely the solution.

MatthewMcGovern
– MatthewMcGovern

09/26/2012 11:35:44
Commented Sep 26, 2012 at 11:35

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Amelia · Accepted Answer · 2012-09-26 11:18:57Z

0

For better formatting you might want to use the heredoc (multi-line string) syntax in PHP.

echo <<<START
    <tr>
    <td> {$row['pid']} </td>
    <td> {$row['Function']} </td>
    <td> {$row['Position']} </td>
    <td> {$row['City']}, {$row['State']} </td>
    <td> {$row['Experience']}</td>
START;

answered Sep 26, 2012 at 11:18

Amelia

2,9702 gold badges26 silver badges39 bronze badges

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