Undefined variables errors - php

Question

So i'm working on a project and i am running to to a few mysql & php errors.
First Error :
This is all code associated with First Error
// Database Class
class Database {
public $_link, $_result, $_numRows;
public function __construct($server, $username, $password, $db){
$this->_link = mysql_connect($server, $username, $password) or die('Cannot Execute:'. mysql_error());
mysql_select_db($db, $this->_link);
}
public function disconnect(){
mysql_close($this->_link);
}
public function query($sql){
$this->_result = mysql_query($sql, $this->_link) or die('Cannot Execute:'. mysql_error());
$this->_numRows = mysql_num_rows($this->_result);
}
public function numRows() {
return $this->_numRows;
}
public function rows(){
$rows = array();
for($x = 0; $x < $this->numRows(); $x++) {
$rows[] = mysql_fetch_assoc($this->_result);
}
return $rows;
}
}
// Setup.php
$is_set = 'false';
global $company_name, $ebay_id;
if( isset($_POST['ebay_id'], $_POST['company_name']) ){
$_ebay_id = $_POST['ebay_id'];
$_company_name = $_POST['company_name'];
$is_set = 'true';
} else {
// Silence is Golden!
}
$_service_database = new Database('localhost', 'root', 'password', 'chat-admin');
$_service_database->query("SELECT * FROM installs WHERE `identity` = '$mysqldb' AND `db_name` = '$mysqldb'");
if ($_service_database->numRows() == 0){
if($is_set == true){
$_service_database->query("INSERT INTO installs (ebay_id, name, db_name, identity) VALUES ('$_ebay_id', '$_company_name', '$mysqldb', '$mysqldb')");
}
} else {
// header("Location: index.php");
}
$_service_database->disconnect();
?>
<form name="setup" method="post" action="<?php echo $_SERVER["PHP_SELF"]; ?>">
<input placeholder="eBay Username" type="text" name="ebay_id">
<input placeholder="Company Name" type="text" name="company_name">
<input type="hidden" name="insert" value="true">
<input type="submit">
</form>
So what this does is checks if the variables defined are in the database if they are not then when the form is submitted it inserts our values, if they are in the database it redirects to index.php, ok so i am receiving the all mighty MySQL num_rows* expects parameter 1 to be resource, boolean given ..., now i have searched everywhere and only found that the die error should solve, that is why you can see it on line 16 in the function query, now although i get the error everything still works its not really good for me to turn error reporting off so any help would be great. Also if you see undefined variables like $mysqldb they are actually set but on a different page and they are valid
Second Error : = function made for simplicity of a templating system
This is all associated with Second Error
function get_content_type($value){
if (strpos($value,'title:') === 0 ) {
$title = explode("title: ",$value);
$value = $title[1];
} else if (strpos($value,'css:') === 0 ) {
$css = explode("css: ",$value);
$value = $css[1];
} else if (strpos($value, 'js:') === 0 ) {
$javascript = explode("js: ", $value);
$value = $javascript[1];
} else {
// Silence is Golden
}
if($value == $title[1]){
echo '<title>'.$value.'</title>';
} else if ($value == $css[1]) {
echo '<link rel="stylesheet" href="'.$value.'">';
} else if ($value == $javascript[1]) {
echo '<script src="'.$value.'"></script>';
}
}
// Calling
get_content_type('title: Welcome to my site');
get_content_type('css: http://something.com/style.css');
get_content_type('js: http://code.jquery.com/jquery-latest.min.js');
everything works as this outputs
<title>Welcome to my site</title><link rel="stylesheet" href="http://somesite.com/style.css"><script src="http://code.jquery.com/jquery-latest.min.js"></script>
Only when i set error_reporting(0);
The errors i get are undefined variables because obviously i'm setting one, then im unsettling that one and setting the other so eventually only 1 is true although its already outputted all of them correctly.
UPDATE
Second Error : Errors =
Notice: Undefined variable: css in C:\Server\www\hidie\libs\test.php on line 19
Notice: Undefined variable: title in C:\Server\www\hidie\libs\test.php on line 17
Notice: Undefined variable: title in C:\Server\www\hidie\libs\test.php on line 17
First Error : Errors =
ERROR: Warning: mysql_num_rows() expects parameter 1 to be resource, boolean given

Answer 1

Ok so here's your problem:
First, as mentioned in the comment, mysql_num_rows only works on select and show. You have it in your query method, which is called by an insert. You can't do that.
public function query($sql){
$this->_result = mysql_query($sql, $this->_link) or die('Cannot Execute:'. mysql_error());
$this->_numRows = mysql_num_rows($this->_result); // <- BAD!
}
You call it on this line:
if($is_set == true){
$_service_database->query("INSERT INTO installs (ebay_id, name, db_name, identity) VALUES ('$_ebay_id', '$_company_name', '$mysqldb', '$mysqldb')");
}
You could make a second method for database and call it for all calls that write to the db:
public function execute($sql){
$this->_result = mysql_query($sql, $this->_link) or die('Cannot Execute:'. mysql_error());
}
which would be called like so:
if($is_set == true){
$_service_database->execute("INSERT INTO installs (ebay_id, name, db_name, identity) VALUES ('$_ebay_id', '$_company_name', '$mysqldb', '$mysqldb')");
}
Second, you need to implicitly define the variables $css and $title outside of those conditionals. Otherwise you are attempting to call undefined vars as indicated by the message...
$title = "";
$css = "";
if (strpos($value,'title:') === 0 ) {
$title = explode("title: ",$value);
$value = $title[1];
} else if (strpos($value,'css:') === 0 ) {
$css = explode("css: ",$value);
$value = $css[1];
} else if (strpos($value, 'js:') === 0 ) {
$javascript = explode("js: ", $value);
$value = $javascript[1];
} else {
// Silence is Golden
}