Cannot get result when passing php variable to mysql query - php

Question

I am trying to pass a part of a file name as a parameter in the my sql query in php. My file name is in the form
db_feature_T1..iOXXdrXXcAMEra_092521.txt
from which I want the part except the db_feature and .txt i.e.
T1..iOXXdrXXcAMEra_092521
I want the img_id from the database whose img_path=dressimages/T1..iOXXdrXXcAMEra_092521.jpg
My code is as follows.I was able to seperate the file name but I cannot get any result from the database.
<?php
include('db.php');
if ($handle = opendir('C:\Users\adithi.a\Desktop\db_features')) {
while (false !== ($entry = readdir($handle))) {
$entry=substr($entry,11,-4); //this seperates the file name db_feature_T1..iOXXdrXXcAMEra_092521.txt to T1..iOXXdrXXcAMEra_092521
$image_path='dressimages/'.$entry.'.jpg';//I want to pass the img_path as it is saved in the database in the form of dressimages/T1..iOXXdrXXcAMEra_092521.jpg
$result= mysql_query("select img_id from tbl_image where img_path='$image_path'") or die("Sorry");
echo $result;//I dont get anything as output.
}
}
closedir($handle);
?>
I went to an infinite loop when executing the code above so i tried:
$image_path='dressimages/'.$entry.'.jpg';//I want to pass the img_path as it is saved in the database in the form of dressimages/T1..iOXXdrXXcAMEra_092521.jpg
$sql = "select img_id from tbl_image where img_path=".$image_path;
echo $sql . '<br />';
$result=mysql_query("$sql");
}
while($data=mysql_fetch_assoc($result))
{
echo $data["img_id"];
}
and Now i get the error mysql_fetch_assoc() expects parameter 1 to be resource, boolean given in C:\wamp\www\FashionBonanza\Readfile.php on line 34
Any help

Answer 1

You have to first fetch the data from your result
$data=mysql_fetch_assoc($result); // this is missing
print_r($data); // or
echo $data["img_id"];
and in case there can be multiple results you can do so in a loop
while($data=mysql_fetch_assoc($result))
{
print_r($data); // or
echo $data["img_id"];
}

Answer 2

Because you're dealing with so many files I would try another approach. That would be to store all paths into an array, make a single query and associate id's with path's. I'm not sure this code will work (I haven't tested it), but I hope you get the picture:
<?php
include('db.php');
if ($handle = opendir('C:\Users\adithi.a\Desktop\db_features')) {
//Store all paths into $image_paths array
$image_paths = array();
while (false !== ($entry = readdir($handle))) {
$entry=substr($entry,11,-4); //this seperates the file name db_feature_T1..iOXXdrXXcAMEra_092521.txt to T1..iOXXdrXXcAMEra_092521
if (strlen($entry) !== 0) {
$image_paths[]='dressimages/'.$entry.'.jpg';//I want to pass the img_path as it is saved in the database in the form of dressimages/T1..iOXXdrXXcAMEra_092521}
}
}
closedir($handle);
//Implode paths
$pathQuotesArray = array_map('apply_quotes', $image_paths); //Looks like 'filename1', 'filename2' etc
$pathQuotes = implode(',', $pathQuotesArray);
//Do one query
$result= mysql_query("select img_id from tbl_image WHERE img_path IN ($pathQuotes)") or die(mysql_error());
//Associate id's with paths
while($data=mysql_fetch_assoc($result))
{
$key[$data["img_id"]] = $data["img_path"];
}
echo $key[5]; //If img_id is 5, then it would show correct path (hopefully :-))
function apply_quotes($item)
{
return "'" . mysql_real_escape_string($item) . "'";
}
?>

Answer 3

Try this
$result= mysql_query("select img_id from tbl_image where img_path="$image_path) or die("Sorry");

Answer 4

Just two ideas:
$result= mysql_query("select img_id from tbl_image where img_path='".$image_path."'") or die("Sorry");
Maybe it's better to do it this way. Otherwise set a extra string variable for the SQL-query, so you can track the exact query, which was executed.
$sQry = "select img_id from tbl_image where img_path='".$image_path."'";
var_dump($sQry);
$result= mysql_query($sQry) or die("Sorry");
So you can check, if the variable contains the correct value or not.
Next point is the "echo". You can't echo a result set, use functions like mysql-fetch-array() or mysql_fetch_assoc() - http://php.net/manual/de/function.mysql-fetch-array.php - or other ones.
while ($row = mysql_fetch_assoc($result)) {
echo $row["img_id "];
}