PHP Echo Function Won't Output into HTML Div with use of innerHTML

Question

Here is the HTML/JS Code:

<script type="text/javascript">
function loadXMLDoc()
{
var xmlhttp;
if (window.XMLHttpRequest)
{// code for IE7+, Firefox, Chrome, Opera, Safari
xmlhttp=new XMLHttpRequest();
}
else
{// code for IE6, IE5
xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
}
xmlhttp.onreadystatechange=function()             {
if (xmlhttp.readyState==4 && xmlhttp.status==200)
{
document.getElementById("dd123").innerHTML=xmlhttp.responseText;
}
}
xmlhttp.open("GET","externalphpcode.php?t=" + Math.random(),true);
xmlhttp.send();
}
window.onload=loadXMLDoc();
</script>

And the External PHP Code:

<?php
header ('Location: urlofHTMLcode');

$con = mysql_connect("hostname","databasename","password");

if (!$con)
{
die('Test Error'.mysql_error());
}

$db_selected = mysql_select_db("databasename", $con);
if (!db_selected)
{
die("Error 3 : ".mysql_error());
}
$result= "SELECT COLUMN_1 FROM T_DEALS WHERE COLUMN_2 = '2011-01-03'
 and COLUMN_3 = 'VALUE_1'" or die ("Error 4 :".mysql_error());

$row = mysql_fetch_array($result);

echo $row;

mysql_close($con);
exit ();

Here's the issue I've been encountering, with much of the HTML/JS redacted for brevity. I've verified that the HTML/JS file is accurately calling the external PHP file by altering the PHP to insert into mySQL. I've been trying to get the SQL Query result to display in a specified DIV container via the PHP echo command paired with the ".responseText" command via AJAX. What am I doing wrong? There doesn't seem to be an issue with the object referencing in HTML, and when debugging the PHP/mySQL connection an error message would replace the DIV with the appropriate error message which is not occurring once the code is altered to call the mySQL data values.

Thank you in advance.

UserZer0 · Accepted Answer · 2011-01-26 07:35:57Z

2

You didn't call mysql_query on your query so your $result might as well have read

 $result="String" Usless statement "String";

Usually you store the sql in a seperate string so you would rewrite as follows:

$sql= "SELECT COLUMN_1 FROM T_DEALS WHERE COLUMN_2 = '2011-01-03' and COLUMN_3 = 'VALUE_1'";
$result= mysql_query($sql);
if(!$result) {die ("Error 4 :".mysql_error());}

Next the $row var is an array which you should either step through with a for loop , or access a specific index. You cant echo arrays directly, all you will ever get is "array" in your output.

The for loop would be something like

 for $cell in $row{
 echo $cell;
 }

and finnally in the code you posted you ommited the final ?> so make sure it's actually there in your php code.

Also I'd really recomend you use jQuery instead of the plan js you are using as it will be less buggy on different browsers, this tutorial will get you started http://docs.jquery.com/Tutorials:Getting_Started_with_jQuery if you haven't heard about it.

edited Jan 26, 2011 at 7:35

answered Jan 25, 2011 at 0:23

UserZer0

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4 Comments

user567623 Over a year ago

I've read on php.net/manual/en/function.mysql-fetch-row.php and w3schools.com/PHP/php_mysql_select.asp that this should work. Even if I call a single value in mySQL it will not echo into the HTML. Maybe the responseText doesn't play nice with the echo function?

UserZer0 Over a year ago

Ok Don't know how I missed this but here's the big one. result didn't make any query. $result should store the "result" of mysql_query(); so change your code to: $result=mysql_query( "SELECT COLUMN_1 FROM T_DEALS WHERE COLUMN_2 = '2011-01-03' and COLUMN_3 = 'VALUE_1'" or die ("Error 4 :".mysql_error()) );

UserZer0 Over a year ago

Also Looking at your other question in context to this, I would again urge you to check out jQuery instead of trying to do the plain Javascript your doing in all your projects. jQuery is a framework for javascript but has much easier ways of doing the things you've been asking about, plus it can do lot's of things you wouldn't think about using plain Js.

user567623 Over a year ago

I'll give that a try once I can make it back to my dev machine. Thank you so much for your input. ...and as for the jQuery business, I've been slow to attempt to learn this because I'm so new at any web development. I'm still trying to get a grasp of js and php at a basic level. But you're probably very right. It's one more thing to learn but from the little I've read it seems very useful.

user567623 · Accepted Answer · 2011-01-29 01:31:37Z

0

To follow up, I have FINALLY found the bug. It seems in my .php call function I put a value in the header which was restricting proper output. It had nothing to do with the .js or the .php language.

UserZer0, you were absolutely right... when I echo $row it outputs array. But I'm finally getting output and can work with that.

A thousand thank you's!

answered Jan 29, 2011 at 1:31

user567623

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