PHP/AJAX/Javascript/MySQL Form Submission

Question

I'm following a tutorial, this version of the form works, where I use onchange. However, the second version below where I try to use onsubmit doesn't work. Can anyone help please?

<form>
<select name="users" onchange="showUser(users.value)">
    <option value="">Select a person:</option>
    <option value="1">Peter Griffin</option>
    <option value="2">Lois Griffin</option>
    <option value="3">Glenn Quagmire</option>
    <option value="4">Joseph Swanson</option>
</select>

This is the one that doesn't work and I'm not sure why?

<form onsubmit="showUser(users.value)">
<select name="users">
    <option value="">Select a person:</option>
    <option value="1">Peter Griffin</option>
    <option value="2">Lois Griffin</option>
    <option value="3">Glenn Quagmire</option>
    <option value="4">Joseph Swanson</option>
</select>
<input type="submit" value="Submit"  >

replace the showUser(users.value) with showUser(this.value)

AMember
– AMember

2012-12-12 11:14:38 +00:00
Commented Dec 12, 2012 at 11:14 — AMember
– AMember, Commented Dec 12, 2012 at 11:14

Rick Kuipers · Accepted Answer · 2012-12-12 11:25:25Z

2

Try this:

<select name="users" onchange="showUser(this[this.selectedIndex].value);">
    <option value="">Select a person:</option>
    <option value="1">Peter Griffin</option>
    <option value="2">Lois Griffin</option>
    <option value="3">Glenn Quagmire</option>
    <option value="4">Joseph Swanson</option>
</select>

For the other one:

<form onsubmit="showUser(this.users[this.users.selectedIndex].value); return false;">
<select name="users">
    <option value="">Select a person:</option>
    <option value="1">Peter Griffin</option>
    <option value="2">Lois Griffin</option>
    <option value="3">Glenn Quagmire</option>
    <option value="4">Joseph Swanson</option>
</select>
<input type="submit" value="Submit" />

Using return false; prevents the browser from posting the form to the url and thus reloading the page.

edited Dec 12, 2012 at 11:25

answered Dec 12, 2012 at 11:12

Rick Kuipers

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1 Comment

user1149620 Over a year ago

Sorry my mistake, I have the code snippets the wrong way around. The one with onchange works, but the one with onsubmit doesn't. Sorry. I am trying to edit the original post.

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