CODEWITHSUNDEEP
phpmysqlsql
Anton
Anton

Reputation: 429

Retrieving Value in PHP

Hello having a bit of problem retrieving a value so i might use it. This is is where i should get my value.

<input type="text" class="typed" readonly="readonly" name="levele"
    value="<?php
    if($get_display->level== "Grade 2") {
        echo "Grade 4"; $lbl="Grade 4";
    }?>">

For a simple test i will use the value Grade 4 because i know what the outcome I'm suppose to have. This is the code that i want to retrieve the value of "levele" so it may perform a SQL statement. For example i have 5 sections that are for grade 4 so i was thinking on using this sql statement.

<select name="section_id">
    <option>SECTIONS</option>
    <?php 
        $result = mysql_query("SELECT * FROM tbl_section where level='".$_GET['levele']."'");
        while($row = mysql_fetch_array($result))
        {
            echo '<option value='.$row['section_id'].'>' .$row['section_name']. '</option>';
        }
    ?>  
</select>

So far i am unable to retrieve "levele" so it may be used it the sql, also tried $_POST and $_request but nothing and i check if levele does have the value Grade 4 and it does also in my database. Would appreciate any help on this. Because after i have retrieve the 5 sections I will try to automatically select one section where for example if the student has a grade of 90 the section should be "A" I'm going to do it with if else. so i would appreciate any help to solve these problems.

Upvotes: 1

Views: 116

Answers (2)

zed Blackbeard
zed Blackbeard

Reputation: 760

<input type="text" class="typed" readonly="readonly" name="levele" value="<?php if( 
 $get_display->level== "Grade 2"){ echo "Grade 4"; **$lbl="Grade 4";**}?>">

remove ** portion and run ....

<select name="section_id">
<option>SECTIONS</option>
<?php 
    $result = mysql_query("SELECT * FROM tbl_section where level='".$_GET['levele']."'");
    while($row = mysql_fetch_array($result))
    {
        echo '<option value='.$row['section_id'].'>' .$row['section`enter code here`_name']. '</option>';
    }
?>  
</select>

Upvotes: 0

Fluffeh
Fluffeh

Reputation: 33542

I see three main problems here that I think you should address:

Firstly, in your code at the top:

<input type="text" class="typed" readonly="readonly" name="levele" value="<?php if( 
     $get_display->level== "Grade 2"){ echo "Grade 4"; $lbl="Grade 4";}?>">

Your code will only echo and set a value if $get_display->level is equal to 2. Did you mean to add in an else statement as well in case the value is something other than Grade 2 perhaps?

<?php 
    if($get_display->level== "Grade 2")
    { 
        echo "Grade 2"; $lbl="Grade 2";
    }
    else
    {
        echo "Grade 4"; $lbl="Grade 4";
    }
?>
<input type="text" class="typed" readonly="readonly" name="levele" value="<?php echo $lbl; ?>">

Secondly, your form section of code seems to be missing quotes around the values:

echo '<option value='.$row['section_id'].'>' .$row['section_name']. '</option>';

I think you meant to use this:

echo '<option value="'.$row['section_id'].'">' .$row['section_name']. '</option>';

Lastly, your code is very vulnerable to injection attacks. You really should look at moving to PDO and prepared statements when dealing with form data.

Edit: This is just an afterthought, but I can't see anywhere that you set $result to a valid connection, but I assume that this is somewhere else in your code.

Upvotes: 2

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