php function returns no result

Question

I wrote a simple php function in an attempt to learn php and functions within php.

The goal of the function is to upload a new user to a db

Here is my function

function upload_user($pId, $pName, $pEmail, $pPhone ){
    $sql = "INSERT INTO users (member_id, name, email, phone) VALUES ('$pId', '$pName', '$pEmail', '$pPhone')";
     mysql_query($sql);
}

Here Is my Form

<form name="register" method="post">
<input type="text" value="name" name="name">
<input type="text" value="email" name="email">
<input type="text" value="phone" name="phone">
<input type="submit" name="submit" value="register" />
</form>

if(isset($_POST['submit'])){
    $id = rand(100, 10000);
    $name = $_POST['name'];
    $email= $_POST['email'];
    $phone = $_POST['phone'];

    upload_user($id,$name,$email,$phone);
}

I know I am connecting successfully to the db since I did a test to echo out if the connection is successful or not. I also have php errors switched onini_set('error_reporting', E_ALL); but it gives me no warnings or errors

My Problem

Nothing happens with the above function, i.e. the result is just blank. If anyone can point out to me what I am doing wrong it will be really appreciated.

Professor Abronsius · Accepted Answer · 2016-02-14 07:26:51Z

I think most likely that the function doen't know about the database connection so you need to either pass the db connection object as a parameter to the function or use the global identifier inside the function.

function upload_user($dbconn, $pId, $pName, $pEmail, $pPhone ){
    $sql = "INSERT INTO `users` (`member_id`, `name`, `email`, `phone`) VALUES ('$pId', '$pName', '$pEmail', '$pPhone')";
     mysql_query($sql,$dbconn);
}

function upload_user($pId, $pName, $pEmail, $pPhone ){
    global $dbconn;
    $sql = "INSERT INTO `users` (`member_id`, `name`, `email`, `phone`) VALUES ('$pId', '$pName', '$pEmail', '$pPhone')";
     mysql_query($sql,$dbconn);
}

Neither of these will return a value - what you return from the function is up to you , if indeed you even want to return a value. You could use the response from the query ( true or false ) as the return:

function upload_user($pId, $pName, $pEmail, $pPhone ){
    global $dbconn;
    $sql = "INSERT INTO `users` (`member_id`, `name`, `email`, `phone`) VALUES ('$pId', '$pName', '$pEmail', '$pPhone')";
     return mysql_query($sql,$dbconn);
}


if(isset($_POST['submit'])){
    $id = rand(100, 10000);
    $name = $_POST['name'];
    $email= $_POST['email'];
    $phone = $_POST['phone'];

    /* using ternary notation to echo something based on return value of function */
    echo upload_user($id,$name,$email,$phone) ? 'Success' : 'Sorry, it failed';
}

You should be aware that certain words, in mysql and other rdbms, have special significance - so it is better to always ( imo ) to encase the field names in backticks. Also worth noting is that the mysql_* suite of functions are deprecated and it s advised to use either mysqli or pdo ~ and where you use user supplied content in your code ( form submission data generally or querystrings ) you should use prepared statements to guard against sql injection.

mehany · Accepted Answer · 2016-02-14 07:35:48Z

0

mysql accepts a link_identifier

link_identifier : The MySQL connection. If the link identifier is not specified, the last link opened by mysql_connect() is assumed. If no such link is found, it will try to create one as if mysql_connect() was called with no arguments. If no connection is found or established, an E_WARNING level error is generated.

reference

So you should pass a connection parameter as noted in other answers. After every connection made, you should close the connection so that you don't run into too manny connections server errors.

answered Feb 14, 2016 at 7:35

mehany

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