php select multiple checkboxes

Question

i have this code, which helps me to retrieve all table fields and associate a check button to them, but this code is generating the same name, i mean it shows me all the fields, but named the same... id

I need their particulatr name..

can you please see what's wrong?

Thanks..

        <form action='report.php' method='post'>

<?php // Script 12.7 - sopping.php

$db = mysql_connect('localhost', 'root', '');
mysql_select_db('db_up', $db);

echo "<table border='1' class='tabtext'>";

$result = mysql_query("SELECT * FROM hostess");
$numrows = mysql_num_rows($result);
$numfields = mysql_num_fields($result);

// show headers
echo '<thead><tr>';
for ($field = 0; $field < $numfields; $field++) {
    $field_name = mysql_field_name($result, $field); // instead of $i
    echo '<th><label><input type="checkbox" name="checkbox[' . $field_name . ']" value="1"/> ' . $field_name . '</label></th>';
}

echo '</tr></thead>';

echo '<tbody>';
for ($row = 0; $row < $numrows; $row++) {
    $data = mysql_fetch_assoc($result);
    echo '<tr>';
    for ($field = 0; $field < $numfields; $field++) {
        $field_name = mysql_field_name($result, $field);
        if (isset($_POST['checkbox'][$field_name])) {
            echo '<td>' . $data[$field_name] . '</td>';
        }
    }
    echo '</tr>';
}
echo '</tbody>';
echo '</table>';


?>
<input type='submit' value='Submit' />
</form>

EmmanuelG · Accepted Answer · 2012-06-18 13:41:23Z

1

The second argument of the mysql_field_name function isn't defined so I am pretty sure PHP is assuming you mean 0 and is returning the first fieldname only. Use the variable you defined ($field) as the index.

Should be:

for ($field = 0; $field < $numfields; $field++) {
    $field_name = mysql_field_name($result, $field); // instead of $i
    echo '<th><label><input type="checkbox" name="checkbox[' . $field_name . ']" value="1"/> ' . $field_name . '</label></th>';
}

answered Jun 18, 2012 at 13:41

EmmanuelG

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9 Comments

Landi Over a year ago

Hey, thanks,, this was obvious, thanks :) Can i please ask you help about one thing, once i select the values, how can display only the selected values? i mean i need to print the fields which i selected, the information isnide them..

EmmanuelG Over a year ago

You already do that but you have the same issue in that block of code as what I answered. If you fixed that, and granted you have the form submitting to this script correctly, it should display only the checked ones. Should be: for ($field = 0; $field < $numfields; $field++) { $field_name = mysql_field_name($result, $field); // not $i again if (isset($_POST['checkbox'][$field_name])) { echo '<td>' . $data[$field_name] . '</td>'; } }

Landi Over a year ago

so i should put inside the report.php this code you just told me?

Landi Over a year ago

Please,s orry for the questions, but is this first part ok? i mean the checking and all stuff? should i proceed with the submission of the form?

EmmanuelG Over a year ago

I am not entirely sure which file your code is in but you should just correct the remaining place that you have this: $field_name = mysql_field_name($result, $i); and replace that $i with $field.

|

Viacheslav Kondratiuk · Accepted Answer · 2012-06-18 13:41:31Z

1

You are using mysql_field_name and it always return the same name for you(for column 0), because $i is not defined.

answered Jun 18, 2012 at 13:41

Viacheslav Kondratiuk

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1 Comment

Landi Over a year ago

Can you please see my updated code? can you help me now with printing all tables , based on the fields i've selected, something like Select $checkbox, $checkbox2 etc ...

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