How to store all the table data in variable in php

Question

My php code is

    $result = mysqli_query($link, "SELECT * FROM album");
if($result)
{
    while ($row = mysqli_fetch_array($result)) 
    {
        $data['path'] = $row['photo_path'];
    }
}
echo json_encode($data);

I want to store the all photos path in $data['path'] What Happening is it taking only last inserted value. I am not getting how can I store the all paths and then echo it.

Answer 1

You need to add array otherwise it will overwrite values

 $result = mysqli_query($link, "SELECT * FROM album");
if($result)
{
    $data = array();
    while ($row = mysqli_fetch_array($result)) 
    {
        $data[]['path'] = $row['photo_path'];
    }
}
echo json_encode($data);

Answer 2

This will also overwrite previous data. So, you are getting last value.

 $data['path'] = $row['photo_path'];

Use, array instead something like

 $data[$i]['path'] = $row['photo_path'];
 $i++; 

Answer 3

If you only need the path you can do:

$result = mysqli_query($link, "SELECT photo_path as path FROM album");
if($result)
{ 
    $data = mysqli_fetch_all($result,MYSQLI_ASSOC);
    echo json_encode($data);         
}

Answer 4

You want to turn the result into an array:

$array = iterator_to_array($result);

The $array is an array that contains all result rows (again as array, array inside array).

More verbose example:

$result = $link->query("SELECT photo_path as path FROM album");
if ($result) {
    $data = iterator_to_array($result);
} else {
    $data = ['error' => $link->error];
}

echo json_encode($data);