mysql query array counter - php

Question

Apologies if I have the terminology wrong.
I have a for loop in php which operates a mysql query...
for ($i = 0; $i <count($user_id_pc); $i++)
{
$query2 = " SELECT job_title, job_info FROM job_description WHERE postcode_ss = '$user_id_pc[$i]'";
$job_data = mysqli_query($dbc, $query2);
$job_results = array();
while ($row = mysqli_fetch_array($job_data))
{
array_push($job_results, $row);
}
}
The results that are given when I insert a...
print_r ($job_results);
On screen -> Array()
If I change the query from $user_id_pc[$i] to $user_id_pc[14] for example I receive one set of results.
If I include this code after the query and inside the for loop
echo $i;
echo $user_id_pc[$i] . "<br>";
I receive the number the counter $i is on followed by the data inside the array for that counter position.
I am not sure why the array $job_results is empty from the query using the counter $i but not if I enter the number manually?
Is it a special character I need to escape?
The full code
<?php
print_r ($user_id_pc);
//Select all columns to see if user has a profile
$query = "SELECT * FROM user_profile WHERE user_id = '" . $_SESSION['user_id'] . "'";
//If the user has an empty profile direct them to the home page
$data = mysqli_query($dbc, $query);
if (mysqli_num_rows($data) == 0)
{
echo '<br><div class="alert alert-warning" role="alert"><h3>Your appear not to be logged on please visit the home page to log on or register. <em>Thank you.</em></h3></div>';
}
//Select data from user and asign them to variables
else
{
$data = mysqli_query($dbc, $query);
if (mysqli_num_rows($data) == 1)
{
$row = mysqli_fetch_array($data);
$cw_job_name = $row['job_description'];
$cw_rate = $row['hourly_rate'];
$job_mileage = $row['mileage'];
$job_postcode = $row['postcode'];
$response_id = $row['user_profile_id'];
}
}
for ($i = 0; $i <count($user_id_pc); $i++)
{
$query2 = " SELECT job_title, job_info FROM job_description WHERE postcode_ss = '{$user_id_pc[$i]}'";
$job_data = mysqli_query($dbc, $query2);
$job_results = array();
while ($row = mysqli_fetch_array($job_data))
{
array_push($job_results, $row);
}
echo $i;
?>
<br>
<?php
}
print ($query2);
print $user_id_pc[$i];
?>

Answer 1

This is primarily a syntax error, the correct syntax should be:
$query2 = " SELECT job_title, job_info FROM job_description WHERE postcode_ss = '{$user_id_pc[$i]}'";
Note that this is correct syntax but still wrong!! For two reasons the first is that it's almost always better (faster, more efficient, takes less resources) to do a join or a subquery or a simple IN(array) type query rather than to loop and query multiple times.
The second issue is that passing parameters in this manner leave your vulnerable to sql injection. You should use prepared statements.
The correct way
if(count($user_id_pc)) {
$stmt = mysqli_stmt_prepare(" SELECT job_title, job_info FROM job_description WHERE postcode_ss = ?");
mysqli_stmt_bind_param($stmt, "s", "'" . implode("','",$user_id_pc) . "'");
mysqli_stmt_execute($stmt);
}
Note that the for loop has been replaced by a simple if

Answer 2

You have to check the query variable, instead of:
$query2 = " SELECT job_title, job_info FROM job_description WHERE postcode_ss = '$user_id_pc[$i]'"
have you tried this:
$query2 = " SELECT job_title, job_info FROM job_description WHERE postcode_ss = '" . $user_id_pc[$i] . "' ";
And another thing, try something different like this:
while ($row = mysqli_fetch_array($job_data))
{
$job_results[] = array("job_title" => $row["job_title"], "job_info" => $row["job_info");
}
Then try to print the values.

Answer 3

Sorry but I like foreach(), so your working code is:
<?php
// To store the result
$job_results = [];
foreach($user_id_pc as $id ){
// selecting matching rows
$query2 ="SELECT job_title, job_info FROM job_description WHERE postcode_ss = '".$id."'";
$job_data = mysqli_query($dbc, $query2);
// checking if query fetch any result
if(mysqli_num_rows($job_data)){
// fetching the result
while ($row = mysqli_fetch_array($job_data)){
// storing resulting row
$job_results[] = $row;
}
}
}
// to print the result
var_dump($job_results);