Why is PHP giving off a undefined variable message for this code?

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?

echo the variable after the $name = $node->nodeValue; to see what the output is — Packet Tracer
– Packet Tracer, Commented Mar 14, 2012 at 19:30

danielrsmith · Accepted Answer · 2012-03-14 19:30:06Z

3

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.

answered Mar 14, 2012 at 19:30

danielrsmith

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Comments

Jon Egeland · Accepted Answer · 2012-03-14 19:30:11Z

1

$node->nodeValue is probably null, which sets $name to null. PHP would render that as undefined.

answered Mar 14, 2012 at 19:30

Jon Egeland

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5 Comments

thetaiko Over a year ago

The if statement right above the assignment would prevent this.

thetaiko Over a year ago

Also, null is certainly not the same as "undefined"

Jon Egeland Over a year ago

@thetaiko that checks nodeName, not nodeValue, and PHPs looseness sometimes renders null as undefined.

thetaiko Over a year ago

my bad on the nodeName/nodeValue. You're correct there. But I will stand by my assertion that PHP will never render null as undefined.

Jon Egeland Over a year ago

@thetaiko I have had a couple cases where this happened. It's rare and depends on whether the variable is an object or not (and more stuff, I don't remember my exact case).

Cerad · Accepted Answer · 2012-03-14 19:33:46Z

1

You left out a set of squiggly brackets

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

    { // *** You left this out 

        $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 

            }

        }
    }
}
}

Also going to need some quotes around $name in your query.

answered Mar 14, 2012 at 19:33

Cerad

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Comments

travega · Accepted Answer · 2012-03-14 19:32:59Z

0

Well not always if $node->nodeName doesn't equal "ModuleName" then the $name variable never gets created. Try defining your $name variable outside your loops or even just outside your if statement.

answered Mar 14, 2012 at 19:32

travega

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Comments

user1248752 · Accepted Answer · 2012-03-14 19:33:31Z

0

$name is out of scope.

Try

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 
                }
            }
        }
    }
}

answered Mar 14, 2012 at 19:33

user1248752

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