php - Inserting Specific Image for MYSQL Database -

i creating application grabs list of instagram images , inserts ones (based on choice) specific database.

currently, when select submit button grabbing last available image in list , inserting database. each image assigned submit button. how make sure proper image, relative submit button, 1 being inserted database.

the following code list of images , submit button each image:

foreach ($media->data $data) { echo $pictureimage = "<img src=\"{$data->images->thumbnail->url}\">"; echo "<form action='tag.php' method='post'>"; echo "<input type='submit' name='submit' value='click me'>";   echo "</form>"; } 

this how inserting image database. remember, grabs last available image in list , inserts that.

if(isset($_post['submit'])) {  // there variables database information $hostname = "random"; $username = "random"; $dbname = "random"; $password = "random!"; $usertable = "random";  //connecting database $con = mysql_connect($hostname, $username, $password) or die ("unable  connect database! please try again later."); mysql_select_db($dbname, $con);  $sql="insert $usertable (image) values ('$pictureimage')"; if (!mysql_query($sql,$con)) {     die('error: ' . mysql_error($con)); }     mysql_close($con); } 

any suggestions?

change form this:

echo "<form action='tag.php' method='post'>"; foreach ($media->data $data) {     echo "<img src=\"{$data->images->thumbnail->url}\">";     echo "<input type=\"hidden\" name=\"image[]\" value=\"".$data->images->thumbnail->url."\">"; } echo "<input type='submit' name='submit' value='click me'>"; echo "</form>"; 

and process $image[] array on backend.

foreach ($_post['image'] $k=>$image){     $sql="insert $usertable (image) values ('$image')";     if (!mysql_query($sql,$con)) {         die('error: ' . mysqli_error($con));     } } 

you might need make few more small changes code logic should work.

update: since op wants 1 button each image, code that:

foreach ($media->data $data) {     echo "<form action='tag.php' method='post'>";     echo "<img src=\"{$data->images->thumbnail->url}\">";     echo "<input type=\"hidden\" name=\"image\" value=\"".$data->images->thumbnail->url."\">";     echo "<input type='submit' name='submit' value='click me'>";     echo "</form>"; } 

the foreach ($_post['image'] ) part should changed to:

$sql="insert $usertable (image) values ('".file_get_contents($_post['image'])."')"; if (!mysql_query($sql,$con)) {     die('error: ' . mysqli_error($con)); } 

this should insert actual image image in database op wants.