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[SOLVED] simple script cannot work

sungpeng

By sungpeng
in PHP Coding Help

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sungpeng Regular Member

sungpeng

Posted
Posted 


<?php

if($_POST[action]=="Submit")
{




include ("".$_SERVER['DOCUMENT_ROOT']."/housing/includes/config.php"); 






$loginname=$_POST[loginname];
echo "$loginname";
$insert_student="insert into users ('llid','loginname') values ('3','$loginname')"; 
mysql_query($insert_student) or die ("cannot insert");

   





?>



<?php }else{ ?>



                 
<?php include ("".$_SERVER['DOCUMENT_ROOT']."/housing/includes/header.php");  ?>

<html>

<head>

<title>Agents Register</title>

</head>

<body>

                 

                     <table width="400" align="center">
                    <br> <form name="form" action="<?php print $PHP_SELF ?>" onSubmit="return validate(this);" method="post">
                      

                         
                        <tr> 
                          <td width="35%">Loginname</td>
                          <td><input type="text" name="loginname" size="25"></td>
                        </tr>
                        
                         

                        <tr> 
                          <td align="center" colspan="2"><input type="submit" name="action" value="Submit"></td>
                        </tr>
                      </table>
                    </form>




          
</body>

</html>
<?php 
}
?>
   

 

Can anyone help, I don't know why simple script why cannot work? always show cannot insert

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dadamssg Advanced Member

dadamssg

Posted
Posted

first...state your problem, what exactly is happening that you don't want...then state what you do want..

 

but try this...you have many unnecessary <?php and ?> tags

 

<?php

if($_POST[action]=="Submit")
{


   

include ("".$_SERVER['DOCUMENT_ROOT']."/housing/includes/config.php"); 
      
$loginname=$_POST[loginname];
echo "$loginname";
$insert_student="insert into users ('llid','loginname') values ('3','$loginname')"; 
mysql_query($insert_student) or die ("cannot insert");

   
}else{ 



                 
include ("".$_SERVER['DOCUMENT_ROOT']."/housing/includes/header.php")}  
?>

<html>

<head>

<title>Agents Register</title>

</head>

<body>

                 

                     <table width="400" align="center">
                    <br> <form name="form" action="<?php print $PHP_SELF ?>" onSubmit="return validate(this);" method="post">
                     

                         
                        <tr>
                          <td width="35%">Loginname</td>
                          <td><input type="text" name="loginname" size="25"></td>
                        </tr>
                       
                         

                        <tr>
                          <td align="center" colspan="2"><input type="submit" name="action" value="Submit"></td>
                        </tr>
                      </table>
                    </form>




         
</body>

</html>

 

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dadamssg Advanced Member

dadamssg

Posted
Posted

and i dont see where you're connecting to your db

 

try this

 

$cxn = mysqli_connect($host,$user,$passwd,$dbname) or die("Couldn't connect");  
    
      
    
   $sql = "INSERT INTO users (llid, loginname) VALUES ('3','{$loginname}')";
   $result = mysqli_query($cxn,$sql)
          or die (mysqli_error($result));

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sungpeng Regular Member

sungpeng

Posted
  • Author
Posted 

$result=mysql_query("insert into users (loginname) values ('$loginname')") or die (mysql_error($result));

 

error is "Warning: mysql_error(): supplied argument is not a valid MySQL-Link resource in signup.php on line 21"  What is it means?

 

 

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sungpeng Regular Member

sungpeng

Posted
  • Author
Posted 

<?php

if($_POST[action]=="Submit")
{




include ("".$_SERVER['DOCUMENT_ROOT']."/housing/includes/config.php"); 




$loginname=$_POST[loginname];

echo "$loginname";


$result=mysql_query("insert into users (loginname) values ('$loginname')") or die (mysqli_error($result)); 


   





}else{ ?>



<html>

<head>

<title>Agents Register</title>

</head>

<body>

                 

                     <table width="400" align="center">
                    <br> <form name="form" action="<?php echo"$PHP_SELF"; ?>" method="post">
                      

                         
                        <tr> 
                          <td width="35%">Loginname</td>
                          <td><input type="text" name="loginname" size="25"></td>
                        </tr>
                        
                         

                        <tr> 
                          <td align="center" colspan="2"><input type="submit" name="action" value="Submit"></td>
                        </tr>
                      </table>
                    </form>




          
</body>

</html>
<?php 
}
?>

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Zane Prolific Member

Zane

Posted
Posted

"Warning: mysql_error(): supplied argument is not a valid MySQL-Link resource in signup.php on line 21" What is it means?

 

It means that you are trying to connect to your database WITHOUT a proper connection variable......

$result = mysqli_query($cxn,$sql)

 

quickest solution: Switch the parameters!

also...you don't need mysqli

 

and you can't call mysql_error() on mysqli connections...if I'm not mistaken....seems pretty logical too

 

 

using mysqli you'd have to connect like this

$cxn = new mysqli("localhost", "my_user", "my_password", "world");

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sungpeng Regular Member

sungpeng

Posted
  • Author
Posted 

<?

$hostname = "localhost";
$username = "sgfai_winning";
$password = "heally";
$dbName = "sgfai_property"; 
MYSQL_CONNECT($hostname, $username, $password) OR DIE("Unable to connect to database"); 
mysql_select_db("$dbName") or die("Unable to select database"); 

?>

 

Hi zanus, is the above connection wrong?

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dadamssg Advanced Member

dadamssg

Posted
Posted

do this

 

$host="localhost";
$user="sgfai_winning";
$passwd = "heally";
$dbname = "sgfai_property";

$cxn = mysqli_connect($host,$user,$passwd,$dbname) or die("Couldn't connect"); 

 

thatll connect you to the right db with the right username, password and you use $cxn in all your mysqli_queries as the first parameter..

like

 

$result = mysqli_query($cxn,$sql)

          or die (mysqli_error($result));

$sql will be your mysql query

 

 

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sungpeng Regular Member

sungpeng

Posted
  • Author
Posted 


<?php

if($_POST[action]=="Submit")
{



$hostname = "localhost";
$username = "sgfai_winning";
$password = "helly";
$dbName = "sgfai_property"; 
MYSQL_CONNECT($hostname, $username, $password) OR DIE("Unable to connect to database"); 
mysql_select_db("$dbName") or die("Unable to select database");


$loginname=$_POST[loginname];

echo "$loginname";

$result=mysql_query("insert into users (loginname) values ('$loginname')") or die (mysqli_error($result)); 



}


?>



<html>

<head>

<title>Agents Register</title>

</head>

<body>

                 

                     <table width="400" align="center">
                    <br> <form name="form" action="<?php echo"$PHP_SELF"; ?>" method="post">
                      

                         
                        <tr> 
                          <td width="35%">Loginname</td>
                          <td><input type="text" name="loginname" size="25"></td>
                        </tr>
                        
                         

                        <tr> 
                          <td align="center" colspan="2"><input type="submit" name="action" value="Submit"></td>
                        </tr>
                      </table>
                    </form>




          
</body>

</html>

 

 

 

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Zane Prolific Member

Zane

Posted
Posted

Now we're gettin somewhere....

 

you've got a table structure problem

apparently you're primary key in your table isn't incrementing...and it isnt even number based....which I don't think is possible, but it may be...never tried.

 

go find your primary key...set it to autoincrement....and check the datatype to make sure that it is INT or TINYINT of something along the lines of INT

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