sending js variables to php page - not working

Question

This is what my javascript looks like:

print "<script>function newpage(){

xmlhttp=new XMLHttpRequest();

var PageToSendTo = 'select_db.php?';
var MyVariable = 'variableData';
var VariablePlaceholder = 'variableName=';
var UrlToSend = PageToSendTo + VariablePlaceholder + MyVariable + '&tweet_id='" .     $id_num;

print "xmlhttp.open('GET', UrlToSend);
xmlhttp.send();
}</script>";

(it is in a php file hence the print statement)

This is what my form submission looks like: (again in a php statement)

print "<form action=\"select_db.php\" onSubmit=\"return newpage()\">";

The function newpage is not working.. I expected the variables to be sent over in the URL but the resulting URL is just select_db.php?

help, please.

Your code has "VariablePlaceholder" twice ... is the second one supposed to be "MyVariable"? It's still not really a valid query string. — Pointy
– Pointy, Commented Jun 26, 2013 at 20:32
You should use the browser console to inspect the http request and see how it's being sent to the server. — Pointy
– Pointy, Commented Jun 26, 2013 at 20:38

mohkhan · Accepted Answer · 2013-06-26 20:35:51Z

1

You are missing the initial <script> tag...

print "<script>function newpage(){

xmlhttp=new XMLHttpRequest();

var PageToSendTo = 'select_db.php?';
var MyVariable = 'variableData';
var VariablePlaceholder = 'variableName=';
var UrlToSend = PageToSendTo + VariablePlaceholder + VariablePlaceholder + '$tweet_id='" .     $id_num;

print "xmlhttp.open('GET', UrlToSend);
xmlhttp.send();
}</script>";

does that solve your problem ?

answered Jun 26, 2013 at 20:35

mohkhan

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2 Comments

user1011332 Over a year ago

It's there, just didn't include it in the code.. will edit now.

mohkhan Over a year ago

Can you say return true; as last line in your function and try ?

paka · Accepted Answer · 2013-06-26 20:30:54Z

0

try this onSubmit=\"newpage(); return false; \"

answered Jun 26, 2013 at 20:30

paka

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2 Comments

Pointy Over a year ago

It won't work to return false before calling the function, because the function will thus never be called.

paka Over a year ago

@Pointy was right; I've update the solution: place function before return

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