Update data from MYSQL, PHP to drop down button

Question

if(isset($_GET["id"])){
        $sql=mysql_query("SELECT * FROM aMovie WHERE aName= '{$_GET['id']}'"); 
        $row=mysql_fetch_object($sql);  
}

<input type = "text" name = "name" value = "<?php echo $row->aC; ?>"/> 
<select name = "name" >
        <option value = "" <?php echo ($row->aC== "Deadpool") ? 'selected = "selected"': '';?>">Deadpool</option>
        <option value = "" <?php echo ($row->aC == "BATMAN VS SUPERMAN") ? 'selected = "selected"': '';?>">BATMAN VS SUPERMAN</option>
</select>

Assume that aMovie is my table name, and in my table there are aName and and aC. However, I would want to display aName which matches aC ["Deadpool" or "Batman Vs Superman"] and display it in the drop down button. It only works for the input type but not the drop down button.

you are putting "selected" inside "value".

Disha V.
– Disha V.

2016-04-12 09:06:17 +00:00
Commented Apr 12, 2016 at 9:06 — Disha V.
– Disha V., Commented Apr 12, 2016 at 9:06

devpro · Accepted Answer · 2016-04-12 09:24:28Z

3

Your <select> should be like:

<select name = "name" >
<option value="Deadpool" <?=($rows->aC == "Deadpool" ? 'selected="selected"': '')?>>Deadpool</option>
<option value="BATMAN VS SUPERMAN" <?=($rows->aC == "BATMAN VS SUPERMAN" ? 'selected="selected"': '')?>>BATMAN VS SUPERMAN</option>
</select>

selected="selected" will use outside the value attribute.

UPDATE:

As @Maninderpreet-Singh mentioned, you also need to change $row to $rows.

edited Apr 12, 2016 at 9:24

answered Apr 12, 2016 at 9:06

devpro

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8 Comments

ikon Over a year ago

Doesn't work, it will still display Deadpool though my data is Batman Vs Superman :(

devpro Over a year ago

@ikon: var_dump($row->aC == "BATMAN VS SUPERMAN") if its return true, than this will work..

ikon Over a year ago

It works for an input box, not the drop down. Drop down still sets Deadpool as the static value

devpro Over a year ago

@ikon: var_dump($row->aC) yout value is not equal to BATMAN VS SUPERMAN

devpro Over a year ago

@ikon: did u checked: var_dump($row->aC == "BATMAN VS SUPERMAN")

|

Ranjeet Singh · Accepted Answer · 2016-04-12 09:05:17Z

1

try to change

<option <?php echo($row->aC== "Deadpool") ? 'selected = "selected"': '';?>  value="<?php echo $row->aC;?>">Deadpool</option>

answered Apr 12, 2016 at 9:05

Ranjeet Singh

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Comments

Maninderpreet Singh · Accepted Answer · 2016-04-12 09:20:31Z

1

try with this and you are using different variable in input

<input type = "text" name = "name" value = "<?php echo $rows->aC; ?>"/>

$row and $rows are different

  <option value = "<?php echo $row->aC; ?>" <?php echo ($row->aC == "Deadpool") ? 'selected':'';?>">Deadpool</option>
  <option value = "<?php echo $row->aC; ?>" <?php echo ($row->aC == "BATMAN VS SUPERMAN") ? 'selected': '';?>">BATMAN VS SUPERMAN</option>

answered Apr 12, 2016 at 9:20

Maninderpreet Singh

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Comments

Maulik · Accepted Answer · 2016-04-12 09:33:04Z

1

<input type = "text" name = "name1" value = "<?php echo $rows->aC; ?>"/> 
<select name = "name2" >
        <option value = " <?php echo($row->aC); ?>" <?php echo($row->aC=="Deadpool")?'selected': '';?>>Deadpool</option>
        <option value = " <?php echo($row->aC);?>" <?php echo($row->aC == "BATMAN VS SUPERMAN")?'selected': '';?>>BATMAN VS SUPERMAN</option>
</select>

edited Apr 12, 2016 at 9:33

answered Apr 12, 2016 at 9:24

Maulik

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1 Comment

Maulik Over a year ago

never take name="name"

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