PHP - Get input from dynamically added html table rows

Question

I have the following Fiddle set up here Fiddle

As you can see, I am able to add inputs by clicking the Add Row button.

All inputs that are added have a unique id and name. The problem is, I cant just do something like

$actionInput = $_POST["actionInput"];

Because I might need

 $actionInput1 = $_POST["actionInput1"]; 
 $actionInput2 = $_POST["actionInput2"]; 
 $actionInput3 = $_POST["actionInput3"];

Depending on how many rows are added. So how can I get all the inputs without knowing what inputs I need to grab?

Thanks

David Ferenczy Rogožan · Accepted Answer · 2016-02-10 11:49:13Z

1

Actually, you need to maintain counter in hidden, which you will get at the time of posting the form in case you don't want to maintain elements as array, otherwise you can put elements as array as described below:

<input type=text name="inputs[]" />

edited Feb 10, 2016 at 11:49

David Ferenczy Rogožan

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answered Feb 10, 2016 at 10:48

Amit Shah

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Comments

Ali Torabi · Accepted Answer · 2016-02-10 10:47:39Z

1

Name your inputs with array boundary, like:

<input type=text name="actioninput[]" />

now you can itreate throught them in you POST or GET ( depends ) array:

print_r($_POST);

answered Feb 10, 2016 at 10:47

Ali Torabi

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Pupil · Accepted Answer · 2016-02-10 10:51:41Z

1

Assuming your JSFiddle works fine for you, following are the steps.

1) Get keys of $_POST.

2) Get maximum counter value from keys.

3) Take a for loop from 0 to count of post.

4) If counter is 0, no suffix, else, add counter as suffix.

5) Now, you get posted variable.

6) Repeat it for every element in rows.

<?php
$keys = array_keys($_POST);
$keys = implode(',', $keys);
$n = str_replace('actionInput', '', $keys);
$m = explode(',', $n);
$max = max($m);

for ($i=0 ; $i<=$max ; $i++) {
    $suffix = ($i==0) ? '' : $i;
    if (isset($_POST['actionInput' . $suffix])) {
        echo "<br/>-".$_POST['actionInput' . $suffix];
    }
}

answered Feb 10, 2016 at 10:51

Pupil

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1 Comment

mickmackusa Over a year ago

This is not what I would recommend. This data is better handled as POST subarrays.

Pupil · Accepted Answer · 2016-02-10 10:58:50Z

1

replace name=actionInput with name=actionInput[] it should be an array with same name same thing will apply with all form fields generated dynamically whose values you'll need.

edited Feb 10, 2016 at 10:58

Pupil

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answered Feb 10, 2016 at 10:48

jarvo69

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Comments

Ajay Sakat · Accepted Answer · 2016-02-10 12:22:04Z

0

Change this line in Add row jquery event return name + i to return name

Remove i from it and then name=actionInput[]

print_r($_POST);

It will gives array of actioninput

edited Feb 10, 2016 at 12:22

answered Feb 10, 2016 at 11:09

Ajay Sakat

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