PHP Submit on Select

Question

I am trying to not have a submit button, is there any way that the form can submit when the user selects a value?

<form method="post" action="<?php echo $_SERVER['PHP_SELF']  ?>" >

    <table class="form">

            <select name="category" class="formfield" id="category">
                <option value="-1"> Category </option>
                <?php
                    $sql_contry = "SELECT * FROM category";
                    $rs_c = mysql_query($sql_contry);
                    while ($row_c = mysql_fetch_array($rs_c)) {
                        echo '<option value="'.$row_c['category'].'">'.$row_c['category'].'</option>';  
                    }
                ?>
             </select>

    </table>

</form>

Use javascript. A similar example is here: http://stackoverflow.com/questions/10021848/enable-disable-submit-button-if-checkbox-is-checked-unchecked — Barış Akkurt, Apr 26 '13 at 19:58

score 29 · Answer 1 · answered Apr 26 '13 at 19:57

29

Here an example

<form>
    <select name='myfield' onchange='this.form.submit()'>
      <option selected>Milk</option>
      <option>Coffee</option>
      <option>Tea</option>
    </select>
    <noscript><input type="submit" value="Submit"></noscript>
</form>

Using noscript tag it allows browsers that are not JavaScript-enabled to still function.

answered Apr 26 '13 at 19:57

TheEwook

11,037
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who the hell uses `noscript` nowdays and why is this relevant here ? – Adidi Apr 26 '13 at 20:08
@Adidi The content inside the – TheEwook Apr 26 '13 at 20:10
ohhhh what do you say ?! I am just saying that in today internet if someone will disable js he could not surf on 90% of the web - the noscript method was relevant to the web 1.0 world... – Adidi Apr 26 '13 at 20:13
I have tried all of the solutions that everyone has posted, and it isn't quite working yet, but I think it is close. Normally, when I select a value and click the submit button, it displays data on the screen. With this method, I can tell that it is submitting, but the data is not being displayed on the screen. Any thoughts? That is, without going through 500 lines of code? ;) – Danny Boy Apr 26 '13 at 21:27
Awsome find! Been Googling around for hours. – Taavi Jan 18 '21 at 20:06

score 4 · Answer 2 · answered Apr 26 '13 at 19:57

4

Yes - use javascript:

Html:

<form id="frm">
   <select onchange="onSelectChange();">
          <option>1</option>
           <option>1</option>
   <select>
</form>

js:

function onSelectChange(){
 document.getElementById('frm').submit();
}

answered Apr 26 '13 at 19:57

Adidi

5,097
4
23
30

1

You actually need to make the js function take an argument or something to make it actually useful. You can't add 2 select lists with this :) . +1 – AKS Apr 26 '13 at 19:59
1

Yeah - just my cents in case OP finds it confusing. – AKS Apr 26 '13 at 20:06

Abid Shah · Answer 3 · 2019-02-20T12:26:05.163

2

<form action="product.php" method="POST">
<select onchange="this.form.submit();" name="prod">
    <option value="">Select product</option>
    <option value="1">abc</option>
    <option value="2">def</option>
    <option value="3">ghi</option>
    <option value="4">jkl</option>
    <option value="5">mno</option>
</select>

edited Feb 20 '19 at 12:26

answered Feb 20 '19 at 11:58

Abid Shah

325
3
5

score 0 · Answer 4 · answered Apr 26 '13 at 20:03

Just change code as below, I haven't check code so there may be some error like capital/small letter I am not sure like, it's submit() or Submit() and so on..

<script language="javascript">
function submitForm(){
    var val = document.myform.category.value;
    if(val!=-1){
        document.myform.submit();
    }
}
</script>
<form method="post" name="myform" action="<?php echo $_SERVER['PHP_SELF']  ?>" >

    <table class="form">

            <select name="category" class="formfield" id="category" onchange="submitForm();">
                <option value="-1"> Category </option>
                <?php
                    $sql_contry = "SELECT * FROM category";
                    $rs_c = mysql_query($sql_contry);
                    while ($row_c = mysql_fetch_array($rs_c)) {
                        echo '<option value="'.$row_c['category'].'">'.$row_c['category'].'</option>';  
                    }
                ?>
             </select>

    </table>

</form>

Thank you, but I copied and pasted the code, and for some reason it does not submit. I looked for capital / small letter errors but could not find any. Any thoughts? — Danny Boy, Apr 27 '13 at 22:37

PHP Submit on Select

4 Answers4