php mysql dropdown issue

Question

I am trying to build a dropdown list from my sql database. Managed to get it work somewhat, but the list only shows the last item in the table, and i briefly get this printed on the page as the page reloads:

[
 {"id":"1","zavod_ime":"PEIT"},
 {"id":"2","zavod_ime":"PTJM"},
 {"id":"3","zavod_ime":"PM"},
 {"id":"4","zavod_ime":"PN"},
 {"id":"5","zavod_ime":"BS"}
]

this is my separate php file which handles connecting and putting things in an array:

    $sqltran = mysqli_query($con, "SELECT id, zavod_ime FROM zavod")or die(mysqli_error($con));
    $arrVal = array();

    while ($rowList = mysqli_fetch_array($sqltran)) {

                    $namez = array(
                        'id'=> $rowList['id'],
                        'zavod_ime'=> $rowList['zavod_ime']
                        );      


                        array_push($arrVal, $namez);    
    }
         echo  json_encode($arrVal);        


    mysqli_close($con);

and the list part in the html:

                    <select>
                        <?php foreach($namez as $zavod): ?>
                        <option value="<?= $namez['id']; ?>"><?= $namez['zavod_ime']; ?></option>
                        <?php endforeach; ?>
                    </select>

ideas?

You're seeing the JSON object as you have echoed a json encoded array. — Mark Railton
– Mark Railton, Commented Jul 18, 2017 at 10:37
change it to this if you want array : echo json_encode($arrVal, true); — Jigar Shah
– Jigar Shah, Commented Jul 18, 2017 at 10:38
you should mark off questions with solutions given; you get good rep back also — Funk Forty Niner
– Funk Forty Niner, Commented Jul 18, 2017 at 11:10

B. Desai · Accepted Answer · 2017-07-18 10:36:41Z

5

You need arrVal to loop instead of namez

<select>
    <?php foreach($arrVal as $namez): ?>
    <option value="<?= $namez['id']; ?>"><?= $namez['zavod_ime']; ?></option>
    <?php endforeach; ?>
</select>

answered Jul 18, 2017 at 10:36

B. Desai

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Comments

Regolith · Accepted Answer · 2017-07-18 10:44:40Z

1

First you are using echo in echo json_encode($arrVal); that is why your data is displaying.
If you are using ajax to retrieve data you should use echo followed by die()

Also you are returning json_encode so before you use that value decode the value.

The following is assuming that you are using include at the top of where you want your select

<?php $arrVal = json_decode($arrVal)?>
<select>
    <?php foreach($arrVal as $namez): ?>
    <option value="<?= $namez['id']; ?>"><?= $namez['zavod_ime']; ?></option>
    <?php endforeach; ?>
</select>

answered Jul 18, 2017 at 10:44

Regolith

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4 Comments

Čedomir Babić Over a year ago

now i am seeing this in the list: Warning: json_decode() expects parameter 1 to be string, array given in /storage/ssd1/720/2097720/public_html/home.php on line 144

Čedomir Babić Over a year ago

and also, how would the code without json_encode go?

Regolith Over a year ago

insted of echo json_encode($arrVal); do this $arrVal = json_encode($arrVal), this will encode the array, but i think now you are getting the data as array so you are getting the error

Čedomir Babić Over a year ago

did that and now i get:Notice: Undefined variable: arrval in /storage/ssd1/720/2097720/public_html/home.php on line 146 Warning: Invalid argument supplied for foreach() in /storage/ssd1/720/2097720/public_html/home.php on line 146

Čedomir Babić · Accepted Answer · 2017-07-18 12:31:59Z

managed to make things work:

    $sqltran = mysqli_query($con, "SELECT id, zavod_ime FROM zavod")or die(mysqli_error($con));
    $arrVal = array();

    while ($rowList = mysqli_fetch_array($sqltran)) {

                    $namez = array(
                        'id'=> $rowList['id'],
                        'zavod_ime'=> $rowList['zavod_ime']
                        );      


                        array_push($arrVal, $namez);    
    }
         echo  json_encode($arrVal);        


    mysqli_close($con);

and html:

                        <?php foreach($arrVal as $namez): ?>
                        <option value="<?= $namez['id']; ?>"><?= $namez['zavod_ime']; ?></option>
                        <?php endforeach; ?>

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