php code not converting to html

Question

I have a php code

<select style="width: 200px;" name="location" id="myselect" onchange="window.location='enable1.php?id='+this.value+'&pos='+this.selectedIndex;">
  <option value="All">All</option>
 <?php

  $sql="select * from location";
  var_dump($sql);
  $query=mysqli_query($conn,$sql);
  echo "1";
  while($row=mysql_fetch_array($query))
  {
      echo "<option value='$row[loc]'>'$row[loc]'</option>";
  }
    ?>

</select>

In this code i echo"options" but instead of that when I view source code I see php part written not options..

This means that the file does not get interpreted as PHP code. So the question is: how do you open it? — arkascha
– arkascha, Commented Jul 1, 2015 at 6:33
And what about this echo "1";? This probably generates invalid HTML, right? — Sebastian Simon
– Sebastian Simon, Commented Jul 1, 2015 at 6:34

Tanvir Ahmad Sohan · Accepted Answer · 2015-07-01 06:52:32Z

2

You used mysql_fetch_array() instead of mysqli_fetch_array()... Try this...

<select style="width: 200px;" name="location" id="myselect" onchange="window.location='enable1.php?id='+this.value+'&pos='+this.selectedIndex;">
<option value="All">All</option>
 <?php

  $sql="select * from location";
  var_dump($sql);
  $query=mysqli_query($conn,$sql);
  echo "1";
  while($row=mysqli_fetch_array($query))
  {
      echo "<option value='$row[loc]'>'$row[loc]'</option>";
  }
    ?>

</select>

let me know if it helps...

answered Jul 1, 2015 at 6:52

Tanvir Ahmad Sohan

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Comments

Pepo_rasta · Accepted Answer · 2015-07-01 06:41:20Z

2

as Saedawke stated in comment, when you see php code instead of output, it means that this file is not interpreted as php but html. make sure that:

you have set file type to php (.php),
your server is running and have php support
your file is not outside of server folder

answered Jul 1, 2015 at 6:41

Pepo_rasta

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1 Comment

Volvox Over a year ago

Add also - "You are running website trough webserver, not by opening file directly in browser" :)

Kapil Verma · Accepted Answer · 2015-07-01 06:34:23Z

0

Try after remove var_dump and echo "1" from your php code hope this helps.

answered Jul 1, 2015 at 6:34

Kapil Verma

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Comments

Alaksandar Jesus Gene · Accepted Answer · 2015-07-01 06:47:11Z

0

Try this. Assumption is you are fetching the data.

<select style="width: 200px;" name="location" id="myselect" onchange="window.location='enable1.php?id='+this.value+'&pos='+this.selectedIndex;">
  <option value="All">All</option>
 <?php

  $sql="select * from location";
  var_dump($sql);
  $query=mysqli_query($conn,$sql);
  echo "1";
  while($row=mysql_fetch_array($query))
  {
      echo "<option value=".$row['loc'].">".$row['loc']."</option>"; //added dots
  }
    ?>

</select>

answered Jul 1, 2015 at 6:47

Alaksandar Jesus Gene

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1 Comment

Sebastian Simon Over a year ago

echo "1"; still needs to be removed.

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