Question

I have this code. It’s just for testing purposes, so you don’t need to tell me to use parameter binding and prepared statements and PDO to avoid SQL Injection.

foreach($dd->getElementsByTagName("ReportItem") as $elmt){
    foreach ($elmt->childNodes as $node){
        if($node->nodeName==="ModuleName")
            $name = $node->nodeValue;

            if($result=mysqli_query($conn,"select * from technology_info where name = $name")){
                if(mysqli_num_rows($result)==0){

                    mysqli_query($conn,"insert into technology_info(id,name,tool_id) values(null,$name,'2')");
                    //ERROR: Undefined variable: name 

                }

            }
        }
}

This is what the code is meant to accomplish: if variable $name is a value that is already in the database, do nothing. Otherwise, add it to the database.

However, I’m getting an error message: Notice: Undefined variable: name in /var/www/teste/index5.php

I mean, the variable is there. Any idea what might be happening?

Answer 1

Because $name is only being set if($node->nodeName==="ModuleName"). That if statement only applies to that one line, yet the code below it (the mysql statements) will continue to run regardless.