PHP form wont submit

Question

I have a simple form:

<form name ="Add" action="Add-Results.php" method="post">
 <label for="name" class="smallfont">Name</label>
 <input type="text" name="name" size="30" />
 <label for="name_error" class="error">Please enter a name</label>
 <input type="image" name="submit" src="../Content/Images/login-submit.gif">
</form>

Which works fine until I add in the following jquery

$(document).ready(function()
{
 $(".error").hide();
 $("input[name=submit]").click(function() 
 {
  $(".error").hide();
  var name = $("input[name=name]").val();

  if (name == "") $("label[for=name_error]").show();

 return false;
 });

});
</script>

When I put in the jquery nothing happens when I click the submit. When I take out the jquery the form posts without a problem. Any ideas?

benhowdle89 · Accepted Answer · 2011-01-31 11:36:26Z

4

if (name == "") $("label[for=name_error]").show();

Should be

if (name == ""){ $("label[for=name_error]").show(); return false}

answered Jan 31, 2011 at 11:36

benhowdle89

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3 Comments

David Mårtensson Over a year ago

@ Mike Muta: To explain the reason for this change, return false will cancel the POST of the form so in your example the onclick method you add always cancel the submit. By wraping the show and return fals inside the if statement you will only cancel submit if you display the error message.

Bazzz Over a year ago

and remove the return false; that is already there and now is always returned and therefore prevents the form to submit.

benhowdle89 Over a year ago

Yup sorry for the lack of explanation but ditto everything @David said

Radek · Accepted Answer · 2011-01-31 11:42:00Z

2

Form will not be sent, becuase you used return false at the end of click handler - so you prevent default action to run. And submitting form is a default action made on click - and it won't be run. Try removing that line.

answered Jan 31, 2011 at 11:42

Radek

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Comments

mdaguerre · Accepted Answer · 2011-01-31 11:53:32Z

0

Maybe you could put all the logic in a function, and on the onclick event of the button call that function, like this:

function checkName() { var flag = true;

$(".error").hide();
var name = $("input[name=name]").val();

if (name == "")
{
    $("label[for=name_error]").show();
    flag = false;
}

return flag;

}

then

<input type="image" name="submit" onClick="checkName()" src="../Content/Images/login-submit.gif">

answered Jan 31, 2011 at 11:53

mdaguerre

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