select menu send value to database

Mostafa · April 9, 2017

Hello

i need update value in database , by choose a value in select menu , i try this code but didn't work



echo "<td>";

                echo "<select name=\"status[]\">"; 

                    echo "<option value=\"open\"> {$row['stu_status']} </option>";

                    echo "<option value=\"close\">closee</option>";

                echo "</select>";

            echo "</td>";

i need the display option ' the one has already insert in database '

and another option

i need 2 option 1- open 2- close

thank you

benanamen · April 9, 2017

It is not updating the database because you have no code to insert into the DB, or did you just forget to show us the relevant code to your post?

That sure is a lot of unnecessary echos. You wouldnt have to escape all the attributes if you used single quotes.

Mostafa · April 9, 2017

yes i have the code

    <?php require_once('../con/config.php'); ?>

<?php
 
    mysql_select_db($database_config, $config);
    mysql_query("set names 'utf8'");

    $query = "SELECT * FROM sec1octa";
    $result = mysql_query($query);
    ?>

<form method="POST" action="edit_data.php">

echo "<table>";
echo "<tr>";
    echo "<th>id</th>";
    echo "<th>name</th>";
    echo "<th>status</th>";
echo "</tr>";
if(mysql_num_rows($result)>0){
    while($row=mysql_fetch_array($result)){
        echo "<tr>";
            echo "<td><input type=\"hidden\" name=\"id[]\" value=\"{$row['stu_no']}\" size=\"15\"></td>";
            echo "<td>{$row['stu_name']}</td>"; 
            echo "<td>";
                echo "<select name=\"status[]\">"; 
                    echo "<option value=\"open\"> {$row['stu_status']} </option>";
                    echo "<option value=\"close\">closee</option>";
                echo "</select>";
            echo "</td>";
        echo "</tr>";
    }
}
?>
</table>
<input class="buttonstyle" type="submit" value="حفظ">

</form>

Mostafa · April 9, 2017

also the update database is work , but i ask about select menu . when i choose option it's work and update database

but the value didn't correct . i can't explain but i wish you can help me

ginerjm · April 9, 2017

Does this actually produce a proper looking html document? Have you looked at the Source in your browser?

If it is correct, then you need to show us the code that does the update!

Mostafa · April 9, 2017

mod.php code

  
    <?php require_once('../Con/config.php'); ?>
   

 <?php

echo '
    <meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<center> ';
    
    
    mysql_select_db($database_config, $config);
    mysql_query("set names 'utf8'");

    $query = "SELECT * FROM sec1octa"; //You don't need a ; like you do in SQL
    $result = mysql_query($query);
    ?>

<form method="POST" action="edit_data.php">

<?php

echo "<table>";
echo "<tr>";
    echo "<th>id</th>";
    echo "<th>name</th>";
    echo "<th>status</th>";
echo "</tr>";
if(mysql_num_rows($result)>0){
    while($row=mysql_fetch_array($result)){
        echo "<tr>";
            echo "<td><input type=\"hidden\" name=\"id[]\" value=\"{$row['stu_no']}\" size=\"15\"></td>";
            echo "<td>{$row['stu_name']}</td>"; 
            echo "<td>";
                echo "<select name=\"status[]\">"; 
                    echo "<option value=\"open\"> {$row['stu_status']} </option>";
                    echo "<option value=\"close\">closee</option>";
                echo "</select>";
            echo "</td>";
        echo "</tr>";
    }
}
?>
</table>
<input class="buttonstyle" type="submit" value="حفظ">

</form>

edit_data.php

    <?php require_once('../Con/config.php'); ?>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
</head>
    <?php

mysql_select_db('gohargro_student');
$db= mysqli_connect("localhost","****","*****","****");


    mysql_query("set names 'utf8'");
       
if(isset($_POST['id'])){
    $tally=0;

    // build all queries for the batch
    foreach($_POST['id'] as $index=>$id){
        $queries[]="UPDATE `goh`.`sec1octa` SET `stu_status`='".mysqli_real_escape_string($db,$_POST['status'][$index])."' WHERE `stu_no`='".mysqli_real_escape_string($db,$id)."'";
    }

    // run all queries
    if(mysqli_multi_query($db,implode(';',$queries))) {
        do{
            $tally+=mysqli_affected_rows($db);
        } while(mysqli_more_results($db) && mysqli_next_result($db));
    }

    // assess the outcome
    if($error_mess=mysqli_error($db)){
        echo "Syntax Error: $error_mess";
    }else{
        echo "$tally row",($tally!=1?"s":"")," updated";
    }
    mysqli_close($con);
}

?>

this is full code , 2 page.

ginerjm · April 9, 2017

If you turned on php error checking you would get some idea of the problems you are having.

You Can't Mix MySQL with mysqli methods. Switch to mysqli and STOP USING THE MYSQL FUNCTIONS.

Mostafa · April 9, 2017

thank you i will use mysqli , but no error happen in select menu that's my main problem now how can i solve it ?

you can check 2 files if you have time and know what i talk about because no error appear .

ginerjm · April 9, 2017

Why don't you echo out all of the queries you are creating and see what they look like?

Mostafa · April 9, 2017

sorry i still understand you , i think you talk about thing i didn't see it ,

i just need to solve the problem about ' select menu ' that's all i need

i need make 2 option in select menu , and insert ( or update ) the choose option in my database

and when i return in page which i choose option form it , appear in select menu the option was already inserted in database and also the other option

Thank you .

ginerjm · April 9, 2017

If the update is not working print out the queries variable to see if the query statements are correct. that is what I would do

It is called "basic debugging".

Did you turn on error checking yet (see my signature)

Mostafa · April 9, 2017

i try but no error appear .

echo "<select name=\"status[]\">"; 
                    echo "<option value=\"open\"> {$row['stu_status']} </option>";
                    echo "<option value=\"close\">closee</option>";
                echo "</select>";

i just need to edit this code , i need select menu have 2 options to insert the chosen options to my database

it's very hard to explain what happen when i click update button it's make update an return value , and if i make another update to another row , the past row also affected .

ginerjm · April 9, 2017

You are showing me your select which looks fine. Why Won't you show me your update code that is not working?????

PS try writing this code like this:

echo "<select name='status[]'>";
echo "<option value='open'>" . $row['stu_status'] . "</option>";
echo "<option value='close'>closee</option>";
echo "</select>";

A lot easier to read.

Jacques1 · April 9, 2017

The code is completely fudged up, and it seems you blindly copied and pasted it from Stackoverflow without understanding how it works.

If you want to be a programmer, you need to understand what you're doing. Writing down random code and relying on people on the Internet doesn't help.

The form you want looks like this:

<!-- Just send the data to the current script; no need for an extra file -->
<form method="post">
    <!-- TODO: add an anti-CSRF token-->
    <input type="hidden" value="<?= html_escape($csrf_token) ?>">
    <table>
        <thead>
            <tr>
                <th>ID</th>
                <th>name</th>
                <th>status</th>
            </tr>
        </thead>
        <tbody>
            <?php foreach ($items as $item): ?>
                <tr>
                    <td><?= html_escape($item['stu_no']) ?></td>
                    <td><?= html_escape($item['stu_name']) ?></td>
                    <td>
                        <select name="status[<?= html_escape($item['stu_no']) ?>]">
                            <option value="open" <?php if ($item['stu_status'] == 'open'): ?>selected<?php endif; ?>>open</option>
                            <option value="closed" <?php if ($item['stu_status'] == 'closed'): ?>selected<?php endif; ?>>closed</option>
                        </select>
                    </td>
                </tr>
            <?php endforeach; ?>
        </tbody>
    </table>
    <input type="submit">
</form>

Do you understand the structure? Each status is stored under the ID of the item. For example, the new status for the item #42 would be $_POST['status']['42']. There's no need for this hidden field stuff.

Depending on the current status in the database, one of the two options is preselected with the selected attribute.

Note that all dynamic input must be HTML-escaped to prevent cross-site scripting attacks, and you need an anti-CSRF token. Also note that if the only possible status values are “open” and “closed”, then you should have a boolean field instead.

Mostafa · April 10, 2017

Thank you , i will try the code and also try to learn it ,

* i make the form but doesn't work with me , that's why i ask here or in stackover to learn but i face some difficult to understand .

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select menu send value to database

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