Pass multiple variables from HTML to PHP

Question

I want to pass two variables to my .php page ... the drop down variable is working great. But when I added an additional variable, it only sends a 0 instead of what I enter in the form.

I feel like I'm fairly close to the solution on this ... when I substitute a number on this line:

    xmlhttp.open("GET","getdeposit.php?q="+str + "&r=" +restaurant, true);

instead of +restaurant, I use + 101 ... then this page correctly passes 101 ... but when I just enter the number in my form, it returns 0

<html>
<head>


<script>
function showUser(str) {
if (str == "") {
    document.getElementById("txtHint").innerHTML = "";
    return;
} else { 
    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("txtHint").innerHTML = xmlhttp.responseText;
        }
    }
    xmlhttp.open("GET","getdeposit.php?q="+str + "&r=" +restaurant, true);
    xmlhttp.send();
}
}
</script>

</head>
<body>

<form>
<p>
 <label for="restaurant">Restaurant:</label>            
 <input type="text" id="restaurant" name="restaurant" placeholder="Enter    Restaurant" />
</p>
<select name="users" onchange="showUser(this.value)">
  <option value="">select a date</optoin>
  <option value=" ">Today</option>
  <option value="1">Yesterday</option>
  <option value="2">Two Days Ago</option>
  </select>
</form>

<div id="txtHint"><b>Select Business Date</b></div>

</body>
</html>

Did you set the variable restaurant in your showUser function? I can't see it, maybe that's the problem. — Sergio B.
– Sergio B., Commented Feb 10, 2015 at 19:20

Endacy · Accepted Answer · 2015-02-10 19:23:29Z

2

Add the following just above the if statement in your js function :

        var restaurant = document.getElementById('restaurant').value;

answered Feb 10, 2015 at 19:23

Endacy

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1 Comment

kloodge Over a year ago

@DeFirmo I added it above the first IF in the js function and it worked for me.

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