Display array list elements in dropdown list in php

Question

For example, I have such array:

$someArray = Array (
    [stationList] => Array (
        [0] => Array (
            [stationName] => A.S.Peta Bypass
            [stationId] => -1 )
        [1] => Array (
            [stationName] => Aala
            [stationId] => -1 )
    )
)

Now U want to display the array elements(stationName) in a dropdown list in php, I used the below code:

<select id="txtLabourId">                                                                                                                          
  <option selected="selected">Choose one</option>

  <?php
    foreach($someArray as $name) { ?>
      <option value="<?php echo $name['stationName'] ?>"><?php echo $name['stationName'] ?></option>
  <?php
    } ?>                                                                                                              
</select>

But its giving the error:

Undefined index stationName on line 222

How to resolve this? Any help appreciated.

check first with stationList

caryarit ferrer
– caryarit ferrer

2017-08-17 08:37:56 +00:00
Commented Aug 17, 2017 at 8:37 — caryarit ferrer
– caryarit ferrer, Commented Aug 17, 2017 at 8:37

RiggsFolly · Accepted Answer · 2017-08-17 08:39:04Z

1

The $someArray array contains another array, so start your foreach loop from the inner array like this

<?php
    foreach($someArray['stationList'] as $station) { 
?>
      <option value="<?php echo $station['stationId'] ?>"><?php echo $station['stationName'] ?></option>
<?php
    } 
?>

answered Aug 17, 2017 at 8:39

RiggsFolly

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Comments

Alex M · Accepted Answer · 2017-08-17 11:02:49Z

1

You should add the inner array stationList into loop.

Your loop code should be like this:

foreach($someArray['stationList'] as $name)

edited Aug 17, 2017 at 11:02

Alex M

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answered Aug 17, 2017 at 8:52

Gautam D

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Comments

Bananaapple · Accepted Answer · 2017-08-17 08:39:46Z

0

You need to give stationList to your foreach, not the wrapping array, so:

foreach($someArray['stationList'] as $name) { ...

answered Aug 17, 2017 at 8:39

Bananaapple

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Comments

Abdallah Salem · Accepted Answer · 2023-02-27 21:02:31Z

0

value type "; echo "$val"; echo "".gettype($val).""; } ?>

answered Feb 27, 2023 at 21:02

Abdallah Salem

1

1 Comment

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