How to display values from database in a checkbox list?

Question

I want to load the data from database to a checkbox list to display and select and here is my code. But it's not working so can anyone correct this code for me?

When i execute this i get the error message as "Parse error: syntax error, unexpected $end in C:\xampp\htdocs\test.php on line 70"

<?php   
    $section = mysql_query("SELECT DISTINCT Section_Text FROM Section");

    while ($row = mysql_fetch_array($section)){
   {
    echo "<tr><td>";
    echo "<input type='checkbox' name='Section' value ='Section_Text'";
    echo " />";
    echo $row['Section_Text'];

    echo "</td></tr><br/>";
   }

?>

ThiefMaster is right, in order to avoid such mistaked in the future , try using a better worksapce - such as eclipse PDT or zend. — user975343
– user975343, Commented Jul 10, 2012 at 21:27
Whats with the <br/> tag ? You echo rows for a table, dont use <br/> divider beetween table rows. And please paste the netire file. Also, use distinct(field) — Ionut Flavius Pogacian
– Ionut Flavius Pogacian, Commented Jul 10, 2012 at 21:29

ThiefMaster · Accepted Answer · 2012-07-10 21:25:51Z

1

You have a { too much right after your while loop. Remove one of them and the code will work.

Besides that, you should really use PDO (or mysqli) instead of the deprecated mysql extension to access your database.

answered Jul 10, 2012 at 21:25

ThiefMaster

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8 Comments

user975343 Over a year ago

Which code line is 70 in your file ? put a note by this codeline , it will be helpful

Anu Over a year ago

I am not getting any error message now but the code is still not working. when i try to display numbers from an array in the checkbox list it works but when i try to display values from database it does not work.

hish Over a year ago

try to run the query alone and see if there is a result

Ionut Flavius Pogacian Over a year ago

in this case, check your database table fields, thats where the error is

Anu Over a year ago

The query is working fine i tried the same by displaying in a dropdown box and its working.

|

ThiefMaster · Accepted Answer · 2012-07-10 21:31:40Z

1

In the while loop :

while ($row = mysql_fetch_array($section)){
   {

Why are there two {? Delete one of them.

edited Jul 10, 2012 at 21:31

ThiefMaster

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answered Jul 10, 2012 at 21:31

hish

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Comments

Fabio Mora · Accepted Answer · 2012-07-10 21:51:33Z

1

Remove the double { after the while declaration.

I suggest you to refactor your code a bit.

Escape your output using htmlentities($row['Section_Text'])
Don't repeat yourself using positional $ printf notation

Use <label ... /> notation to keep your form accessible: check this

<?php   
$section = mysql_query("SELECT DISTINCT Section_Text FROM Section");

$n = 0;
while ($row = mysql_fetch_array($section))
{
    print "<tr><td>" . PHP_EOL;
    printf('<input type="checkbox" name="section_%2$d" id="section_%2$d" value="%1$s" />
         <label for="section_%2$d">%1$s</label>' . PHP_EOL,
          htmlentities($row['Section_Text'], ENT_QUOTES), $n++
    );
    print "</td></tr>" . PHP_EOL;
}

answered Jul 10, 2012 at 21:51

Fabio Mora

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Comments

3rikster · Accepted Answer · 2012-07-10 21:34:36Z

0

there are 2 { in the while, remove one. and also get rid of the
tags, use css to get the desired padding

answered Jul 10, 2012 at 21:34

3rikster

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