Quote:
You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'll find a job description example for most common jobs. Starting with a sampl' at line 1
Because of the way you build the query, its syntax depend on the contain of variables used to built it, we have no way to know what is wrong.
For debugging purpose, add
echo $sql;
in your code to know what is the exact query.
Never build an SQL query by concatenating strings. Sooner or later, you will do it with user inputs, and this opens door to a vulnerability named "SQL injection", it is dangerous for your database and error prone.
A single quote in a name and your program crash. If a user input a name like "Brian O'Conner" can crash your app, it is an SQL injection vulnerability, and the crash is the least of the problems, a malicious user input and it is promoted to SQL commands with all credentials.
SQL injection - Wikipedia[
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SQL Injection[
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SQL Injection Attacks by Example[
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PHP: SQL Injection - Manual[
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SQL Injection Prevention Cheat Sheet - OWASP[
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How can I explain SQL injection without technical jargon? - Information Security Stack Exchange[
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