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I have looked around and Im still not sure on how to do this. I have dried several different ways and its obvious im still doing it wrong. Please can you help

I have an accounts table containing the account username and password and a separate contacts table which is liked to the accounts table by the username. I need to insert in to both of these. here is what I have so far.

Thanks

//signup.php  
include 'connect.php';  

echo '<h3>Sign up</h3>';  

if($_SERVER['REQUEST_METHOD'] != 'POST')  
{  
    /*the form hasn't been posted yet, display it 
      note that the action="" will cause the form to post to the same page it is on */  
    echo '<form action="" method="post">
  <br>
  <table width="0" border="0">
    <tr>
    <th align="left" scope="col">Name:</th>
    <th scope="col"><input type="text" name="name"></th>
  </tr>
  <tr>
    <th align="left" scope="row">Phone:</th>
    <td><input type="text" name="phone"></td>
  </tr>
  <tr>
    <th align="left" scope="row">Address</th>
    <td><textarea name="address" rows="4"></textarea></td>
  </tr>
  <tr>
    <th align="left" scope="row"><p>Postcode</p></th>
    <th align="left" scope="row"><input type="text" name="postcode" id="postcode"></th>
  </tr>
  <tr>
    <th align="left" scope="row">Email</th>
    <td><input type="text" name="email"></td>
  </tr>
  <tr>
    <th align="left" scope="row">Username</th>
    <td><input type="type" name="username"></td>
  </tr>
  <tr>
    <th align="left" scope="row">Password</th>
    <td align="left"><input type="password" name="password"></td>
  </tr>
  <tr align="left">
    <th colspan="2" scope="row"><input type="Submit"></th>
  </tr>
  </table>
</form>'; 
} 
else 
{ 
    /* so, the form has been posted, we'll process the data in three steps:  
        1.  Check the data  
        2.  Let the user refill the wrong fields (if necessary)  
        3.  Save the data  
    */  
    $errors = array(); /* declare the array for later use */  

    if(isset($_POST['username']))  
    {  
        //the user name exists  
        if(!ctype_alnum($_POST['username']))  
        {  
            $errors[] = 'The username can only contain letters and digits.';  
        }  
        if(strlen($_POST['username']) > 30)  
        {  
            $errors[] = 'The username cannot be longer than 30 characters.';  
        }  
    }  
    else  
    {  
        $errors[] = 'The username field must not be empty.';  
    } 
    if(!empty($errors)) /*check for an empty array, if there are errors, they're in this array (note the ! operator)*/  
    {  
        echo 'Uh-oh.. a couple of fields are not filled in correctly..'; 
        echo '<ul>'; 
        foreach($errors as $key => $value) /* walk through the array so all the errors get displayed */ 
        { 
            echo '<li>' . $value . '</li>'; /* this generates a nice error list */ 
        } 
        echo '</ul>'; 
    } 
    else 
    { 
        //the form has been posted without, so save it 
        //notice the use of mysql_real_escape_string, keep everything safe! 
        //also notice the sha1 function which hashes the password 
        $sql = "INSERT INTO 
                    tbl_accounts(accounts_username, accounts_password, accounts_date) 
                VALUES('" . mysql_real_escape_string($_POST['username']) . "', 
                       '" . sha1($_POST['password']) . "', 
                        NOW())";  
        $sql2= "INSERT INTO 
                    tbl_contacts(contacts_username, contacts_name, contacts_email, contacts_phone, contacts_address, contacts_postcode, contacts_date) 
                VALUES('" . mysql_real_escape_string($_POST['username']) . "',
                '" . mysql_real_escape_string($_POST['name']) . "',
                '" . mysql_real_escape_string($_POST['email']) . "',
                '" . mysql_real_escape_string($_POST['phone']) . "',
                '" . mysql_real_escape_string($_POST['address']) . "',
                '" . mysql_real_escape_string($_POST['postcode']) . "',
                        NOW())";

        $result = mysql_query($sql);  
        if(!$result)  
        $result = mysql_query($sql2);  
        if(!$result)  
        {  
            //something went wrong, display the error  
            echo 'Something went wrong while registering. Please try again later.'; 
            //echo mysql_error(); //debugging purposes, uncomment when needed 
        } 
        else 
        { 
            echo 'Successfully registered. You can now <a href="signin.php">sign in</a> and start posting! :-)'; 
        } 

    } 
}
ilanco
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Chris Wilson
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3 Answers3

1

Change:

    $result = mysql_query($sql);  
    if(!$result)  
    $result = mysql_query($sql2);  
    if(!$result)  
    {  

To:

    if(!mysql_query($sql) || !mysql_query($sql2))  
    {  

...and I think your problem will disappear.

It would be worth you re-reading this and this.

Note also that you should be using the POST/Redirect/GET design pattern for handling form submissions of this nature.

DaveRandom
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  • Your answer cleans up his problem much better than mine does. – Jake May 22 '12 at 21:31
  • I have tried this and it still doesnt work.-It adds it to the accounts database but not the contacts:/ – Chris Wilson May 22 '12 at 21:54
  • @ChrisWilson Do you get your `Something went wrong while registering. Please try again later.` message? What error message do you see from `mysql_error()`? – DaveRandom May 22 '12 at 22:01
  • I changed the query to the one below and its the second query that is failing... (the inset in to contacts) it just said second query failed – Chris Wilson May 23 '12 at 13:18
0

This code block doesn't do what you think it does

What this does is sends query sql1 and then if that query failed does sql2.

    $result = mysql_query($sql);  
    if(!$result)  
    $result = mysql_query($sql2);  
    if(!$result)  

If you want to insert into both of them then you should do

    $result = mysql_query($sql);  
    if(!$result)  {
       echo "first query failed"; 
    }
    else {  $result = mysql_query($sql2); }  

    if(!$result) {
       echo "second query failed";
     }
Jake
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  • Keeps saying second query failed. It is becuase it doesnt like two things been added in but i dont know how to do this correctly – Chris Wilson May 22 '12 at 21:58
  • @ChrisWilson if it says the second query is failing that means that your second query is not legal. I feel like that error message is pretty self explanatory. Look into phpmyadmin or toad. – Jake May 30 '12 at 23:43
0

Searching stackoverflow you'll find links like those below:

Stack 1

Stack 2

Community
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djot
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