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Editorial Team
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Asked: 2026-06-15T14:10:31+00:00

I have a form, example: <form method=post action=some external URL> <input type=hidden id=id name=session_id

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I have a form, example:

<form method="post" action="some external URL">
    <input type="hidden" id="id" name="session_id" />
    <input type="hidden" id="val1" name="val1"  />
    <input type="hidden" id="val2" name="val2" />
    <input type="submit" value="Send" />
</form>

I need to save this data into mySQL before sending it into “some external url” – let’s say that my “saving to database” script is on “script/save” url. How can I transfer my data there before sending it throught this form?

EDIT

It would be nice, when my form will only be sent to “some external url” when the “script/save” gives POSITIVE RESPONSE.
Could you help me with that too?

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1 Answer

  1. Editorial Team

    Hey just use an ajax call onsubmit then submit the form when the ajax call is done

    <html>
<head>
<script type="text/javascript">
    function doSomething()
    {
        xmlHttp = new XMLHttpRequest();
        xmlHttp.open('GET', 'yourscript.php, true);
        xmlHttp.onreadystatechange = function(){
            if (xmlHttp.readyState == 4 && xmlHttp.status == 200){
                var response = xmlHttp.responseText;
                // response is what you return from the script
                if(response == 1){
                    alert('works');
                    form[0].submit;
                }else{
                    alert('fails');
                }
            }else{
                alert('fails');
            }
        }
        xmlHttp.send(null);
    }
</script>
</head>
<body>
    <form action="external-url" method="post" onsubmit="doSomething();">
        <input type="text" name="blah" id="blah">
        <input type="submit" value="save">
    </form>
</body>
</html>
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