My first complex function

[V]

I have been using several functions for each query but now I'm attempting to create just one function that queries from the database by using 2 arguments. 1 argument for the DB table and another for the column. Perhaps I need more experience to do this but I'm really curious if I'm somewhat close.

 

 

function listQuery($table,$column) {

//1. Create DB connection

$connection = mysql_connect("localhost", "user", "password"); 
if (!$connection) {
	die("Database connection failed: " .mysql_error());
}

//2. Select a DB to use

$db_select = mysql_select_db("sitetest",$connection);
if (!$db_select) {
	die("Database selection failed: " .mysql_error());
}

$table = '';
$column = '';

return 

mysql_query("SELECT * FROM $table", $connection);
if (!$table) {
	die("Database query failed: " .mysql_error());
} 

while ($row = mysql_fetch_array($table)) { 

	echo $row['$column'];
}
}

 

and I'm using this code to echo the loop

 

echo listQuery('categories','cat_name');

 

My idea was to list all my categories from the DB but instead I get a blank page. No errors tho.

 

What are your thoughts?

[joel24]

you're wiping the content of the column variable in the line:

$column = '';

 

and also you're just executing the MYSQL query but not storing the results in a variable, you need to have @variable = mysql_query("select * from rararara");

 

and your returning nothing from the function.

 

if you want to return an array of the db content instead of echoing it, you can just have

return mysql_fetch_array($sql)

instead of the while loop.

try

 

function listQuery($table,$column) {






//1. Create DB connection






$connection = mysql_connect("localhost", "user", "password"); 



if (!$connection) {





die("Database connection failed: " .mysql_error());



}



//2. Select a DB to use



$db_select = mysql_select_db("sitetest",$connection);



if (!$db_select) {





die("Database selection failed: " .mysql_error());



}



$sql = @mysql_query("SELECT * FROM $table", $connection);



if (!$sql) {





die("Database query failed: " .mysql_error());



} 






while ($row = mysql_fetch_array($sql)) { 








echo $row['$column'];
}
}

[V]

Joel that's very cool, thank you! I'm able to query using the $table variable but I get

 

Notice: Undefined index: $column in C:\wamp\www\gisttest\functions.php on line 26

 

Notice: Undefined index: $column in C:\wamp\www\gisttest\functions.php on line 26

 

Notice: Undefined index: $column in C:\wamp\www\gisttest\functions.php on line 26

 

Notice: Undefined index: $column in C:\wamp\www\gisttest\functions.php on line 26

 

 

for the $column. It works however when I type the column name like this

 

echo $row['cat_name']; instead of $column. Would you happen to know why the error appears?

 

 

 

 

[KevinM1]

Joel that's very cool, thank you! I'm able to query using the $table variable but I get

 

Notice: Undefined index: $column in C:\wamp\www\gisttest\functions.php on line 26

 

Notice: Undefined index: $column in C:\wamp\www\gisttest\functions.php on line 26

 

Notice: Undefined index: $column in C:\wamp\www\gisttest\functions.php on line 26

 

Notice: Undefined index: $column in C:\wamp\www\gisttest\functions.php on line 26

 

 

for the $column. It works however when I type the column name like this

 

echo $row['cat_name']; instead of $column. Would you happen to know why the error appears?

 

 

 

 

 

You assigned '' to $column, so when you use $row['$column'], you're actually attempting to access $row[''], which obviously doesn't exist.  In order to access a table row column, you need to either know the column name, or use a foreach to cycle through them all.

 

foreach ($row as $key => $value)
{
   echo "Column name: $key -- Value: $value";
}

[V]

Oh I see. I guess I'll just type the column for now.