Debug var on php inside js

Question

I have a script (editablegrid) that loads an php file (loaddata.php) inside a js file(JSON)(demo.js). When i make a js alert on php file:

echo '<script language="javascript">';
echo 'alert("message successfully sent")';
echo '</script>';

Script stopped and alert it's not show. If i remove this echo's script works fine. How can i make an alert to see a var (an $sql var)?

Thanks

According to your demo url, you are loading it like this : this.editableGrid.loadJSON(url); ? If yes, try to load it using different way since the data is supposed to be json — w00d
– w00d, Commented Mar 7, 2015 at 17:32
I can do with XML, but with loadXML, i can't do any echo on loaddata.php neither — JavierCabot
– JavierCabot, Commented Mar 8, 2015 at 16:54

Alvaro Flaño Larrondo · Accepted Answer · 2015-03-07 17:28:19Z

1

I suspect that you are echoing inside an already opened <script> tag.

Try removing echo '<script language="javascript">'; and echo '</script>'.

answered Mar 7, 2015 at 17:28

Alvaro Flaño Larrondo

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5 Comments

JavierCabot Over a year ago

No, it doesnt works. If i write only echo someting script stop and not show data. If i remove this echo data is showed well. And i need to write echo for debug a var. Thanks

Alvaro Flaño Larrondo Over a year ago

Probably the console is throwing some info that will help us know what is going on. Press F12, select the console and paste here what it is showing.

JavierCabot Over a year ago

Here it's error SyntaxError: missing ) in parenthetical (test{"metadata":[{"name":"post_title","label":"Cliente","datatyp. Test it's a echo

Alvaro Flaño Larrondo Over a year ago

Looks like the JSON data is badly formatted. Probably the closing ) is missing.

JavierCabot Over a year ago

But this error it's only when i make an "echo ,,,,". I think that when encode JSON, it's encoded badly and shows this error. But i don't know how can i do it

Vladimir Ramik · Accepted Answer · 2015-03-07 17:18:30Z

0

To debug I use the following to save data from php into javascript. Flavor to your liking:

<?php
$aTmp[ 'data1' ] = 'coolstuff1!';
$aTmp[ 'data2' ] = 'coolstuff2!';
echo '<script>';
echo "var oData = [];";
echo 'var oData = ' . json_encode( $aTmp ) . ';';
echo 'console.log( oData );';
echo '</script>';
?>

edited Mar 7, 2015 at 17:18

answered Mar 7, 2015 at 16:39

Vladimir Ramik

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11 Comments

JavierCabot Over a year ago

If i paste this code, script stopped and message it's not show. I don't know why. If i put a simple echo "Hello", script don't run

Vladimir Ramik Over a year ago

Use browser's console to inspect data. Right click open console and you'll see the data under logging section. Check out firebug! addons.mozilla.org/en-us/firefox/addon/firebug

JavierCabot Over a year ago

I can't see on Firebug. Look gestoriabarcelona.com/gestion/demo/…

JavierCabot Over a year ago

Can anyone install editablegrid an test how to make a debug inside loaddata.php file?

Vladimir Ramik Over a year ago

I see a syntax error on firefox: Right click, Inspect Element, SyntaxError: expected expression, got '<' Can you post the file you're running? I edit my original post to include php tags you may have been missing.

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