Compare values from array mysql php [duplicate] - php

Question

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Passing an array to a query using a WHERE clause
(17 answers)
Closed 1 year ago.
Hi I'm trying to compare values from 2 arrays with this query, which is the only way I know:
$session = "1,2,3,"
$table_name = "table1";
$column_name = "data1"; // value for test is 1,4,5,
$sql = "";
$sql .= "SELECT * FROM $table_name WHERE ";
$franquia = array();
$franquia = explode(",", $session);
if (!empty($franquia)) {
$final_id = array();
foreach ($franquia as $val) {
if (trim($val) != '') {
$final_id[] = $val;
}
}
$count_data = count($final_id);
foreach ($final_id as $key => $id) {
if ($id > 0) {
$sql .= " $id IN ($column_name) ";
if ($key < $count_data - 1) {
$sql .= "OR ";
}
}
}
}
echo $sql;
I have values 1,2,3 on $session and 1,4,5on $data1 so the comparison between $session and $data1 was suposed to return true due to both of them have value 1, but I don't get any results.
Actually it only works if both arrays are the same, like $session = 1,2,3 and $data1 = 1,2,3
What am I doing wrong?

Answer 1

You are using the IN clause wrong; In a nutshell, instead of
WHERE value IN (column)
The correct usage is:
WHERE column IN (value, value, value)
Which in turn will not help with what you are trying to do (more about the IN clause). Rather, try the following:
foreach ($final_id as $key => $id) {
if ($id > 0) {
$sql .= "$column_name LIKE %$id,%";
if ($key < $count_data - 1) {
$sql .= "OR ";
}
}
}
}
This should work, but there are a couple things you should be aware of. First you probably are vulnerable to SQL Injections, and second you should consider either using a different scheme with your data in your DB, since using LIKE %% to search for numeric values is a waste, or fetch in advance the entry from the DB and search in the resulting string. Consider the code above nothing but a quick and dirty hack that you should rather avoid :)