PHP - How to update multiple row using foreach?

Question

so i'm making a php file that updates multiple rows to mysql but I'm having a problem whenever i submit,and I have no idea if I'm using foreach well. here is my code:

$query = "SELECT id, departments, deptName, headOfOffice FROM aip";
$result = mysqli_query($db,$query);
$count = mysqli_num_rows($result);

while ($row = mysqli_fetch_assoc($result)) {

        echo '<tr>';
        echo '<td><input type="text" name="id[]" value="'.$row['id'].'" readonly></td>';
        echo '<td><input type="text" id ="department_code" name="department_code[]" value="'.$row['departments'].'"></td>';
        echo '<td><input type="text" id="department_name" name="department_name[]" value="'.$row['deptName'].'"></td>';
        echo '<td><input type="text" id="department_head" name="department_head[]" value="'.$row['headOfOffice'].'"></td>';
        echo '</tr>';

}

echo '<tr>';
echo '<td></td>';
echo '<td></td>';
echo '<td><input type="submit" name="update" value="Update">';
echo '</tr>';



if($_SERVER["REQUEST_METHOD"] == "POST"){

$deptid = $_POST['id'];
$code = $_POST['department_code'];
$dname = $_POST['department_name'];
$dhead =$_POST['department_head'];

foreach($_POST['id'] as $count){ \\ i don't know if this is right.

$query2 = "UPDATE aip SET deparments = '".$code[$count].'" WHERE id = "'.$deptid[$count]."'";
$result2 = mysqli_query($db,$query2);
}

 }

the error says "Undefined offset: 2" I'm a newbie here, and this is my first time using arrays. hope someone could help. please!

You should use prepared statements, which make errors like this almost not possible. Also, you might be able to do a single UPDATE instead of looping. — Tim Biegeleisen
– Tim Biegeleisen, Commented Feb 9, 2017 at 9:14

shukshin.ivan · Accepted Answer · 2017-02-09 09:22:35Z

3

foreach($_POST['id'] as $count => $id){
    $query2 = "UPDATE aip SET deparments = '".$code[$count]."' WHERE id = '".$deptid[$count]."'";
    $result2 = mysqli_query($db,$query2);
}

P.S. your code is vulnerable to SQL injection

edited Feb 9, 2017 at 9:22

answered Feb 9, 2017 at 9:15

shukshin.ivan

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2 Comments

Marky Over a year ago

i already made the changes, the errors are gone but still it doesn't update.

shukshin.ivan Over a year ago

place updating code at the start of a file otherwise you don't see changes. Try to refresh page to see your changes.

Luka Govedič · Accepted Answer · 2017-02-09 09:18:41Z

0

There are two ways of using foreach:

foreach($array as $key => $value){
    ...
}

or

foreach($array as $value){
    ...
}

The foreach loop will then iterate through all elements in array, executing the code in brackets for each element once, holding the key name and the value of the element in $key and $value respectively (you can use other than $value and $key as well).

I don't fully understand what exactly you are doing in your code, so if you add an explanation, I will be able to solve your particular case;

answered Feb 9, 2017 at 9:18

Luka Govedič

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Comments

Dilip kumar · Accepted Answer · 2017-02-09 09:34:20Z

0

i have changed in your script..pls used my script. you do not need to changes any thing just copy and paste my script.

    if($_SERVER["REQUEST_METHOD"] == "POST"){
         $data = $_POST;

        if(count($data['id'])>0){
          foreach($data as $key=>$item){
              $query2 = "UPDATE aip SET deparments = '".$item['department_code'][$key]."', deptName = '".$item['department_name'][$key]."',headOfOffice = '".$item['department_head'][$key]."'  WHERE id = '".$item['id'][$key]."'";
              $result2 = mysqli_query($db,$query2);
        }
   } 
}

answered Feb 9, 2017 at 9:34

Dilip kumar

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