check if number already exists in database

saltedm8 · September 7, 2009

I have a form that I am submitting to the database, one of those fields is id, that has to be a number, I need to be able to submit the number ( within the form choices user just types in any number they want ) and check to see if the number already exists in the database, if it exists I want to display a message to choose another id number

I am having trouble comparing the numbers in the database, how can I check the number and test to see if there are any of the same number already in the database ?

<?php 
include("connect.php");

$id = $_POST['id'];
$name	= $_POST['name'];
$address	= $_POST['address'];
$email	= $_POST['email'];

if($_POST['submit']){

$turn = mysql_query("SELECT letable WHERE id ='$id'");
if ($turn = $id){

	echo 'that number exists already';

	}

if($id){

	mysql_query("INSERT INTO letable (`id`,`name`,`address`,`email`)VALUES('$id','$name','$address','$email')");

}if(!$id){

	echo "<strong>id has not been filled in</strong>";

}

              }



?>

as you can see, I have tried, but obviously in that case id is the same as turn anyway lol

thank you

kratsg · September 7, 2009

try

$id = (int)$_POST['id'];

Make sure in the database, the column storing the number is set to some type of "...INT"

mikesta707 · September 7, 2009

one, turn is a mysql resource, not the results itself. You have to use mysql_fetch_array or one of the other functions that fetches the data.

also, if id is an integer column, don't surround what you are comparing it to with quotes. that denotes it as a string, which may cause problems

also, since you are testing if a row is present with the id, you don't need to compare the values, since they are already the same. you should check if the query returned any results

try

$turn = mysql_query("SELECT letable WHERE id =$id'");
$num = mysql_num_rows($turn);
if ($num > 0){
//no

Edit: one last thing. what you have here

if ($turn = $id){

is valid syntax. but incorrect logic. You are using the assignment operator (=) which assigns the value of the right operand to the left. You want to use the comparison operator(==) which will compare two values

if ($turn == $id){

very simple mistake that is commonly made, but just watch out for it

kratsg · September 7, 2009






if(mysql_num_rows($turn = mysql_query("SELECT letable WHERE id =$id LIMIT 1")))
echo 'that number exists already';

Try that instead.

mikesta707 · September 7, 2009

if(mysql_num_rows(mysql_query("SELECT letable WHERE id =$id LIMIT 1")))

shorter than my example, but basically the same thing

saltedm8 · September 7, 2009

I should have warned you I am a beginner :shrug:

this is what I changed it to

<?php 
include("connect.php");

$id = (int)$_POST['id'];
$name	= $_POST['name'];
$address	= $_POST['address'];
$email	= $_POST['email'];


if($_POST['submit']){


if(mysql_num_rows($turn = mysql_query("SELECT letable WHERE id =$id LIMIT 1")))
       echo 'that number exists already';

if($id){

	mysql_query("INSERT INTO letable (`id`,`name`,`address`,`email`)VALUES('$id','$name','$address','$email')");

}if(!$id){

	echo "<strong>id has not been filled in</strong>";

}

              }



?>

and this was the result after submission and purposely putting in an existing number ( and it still posted in the database )

Warning: mysql_num_rows() expects parameter 1 to be resource, boolean given in G:\wamp\www\lessons\homeworkdynamicform.php on line 36

mikesta707 · September 7, 2009

thats because,



if(mysql_num_rows($turn = mysql_query("SELECT letable WHERE id =$id LIMIT 1")))

is incorrect syntax.

$turn = mysql_query("SELECT letable WHERE id =$id LIMIT 1")

is a boolean statement. It basically means if $turn is assigned the value of that query, return true. else return false

I posted the fixed version one post up. change it to that

kratsg · September 7, 2009

thats because,
if(mysql_num_rows($turn = mysql_query("SELECT letable WHERE id =$id LIMIT 1")))
is incorrect syntax.
$turn = mysql_query("SELECT letable WHERE id =$id LIMIT 1")
is a boolean statement. It basically means if $turn is assigned the value of that query, return true. else return false

I posted the fixed version one post up. change it to that

Actually, they didn't have the right syntax :-o Notice how there's no "FROM table" in there.

if(mysql_num_rows($turn = mysql_query("SELECT * FROM letable WHERE id =$id LIMIT 1")))

saltedm8 · September 7, 2009

thank you, that last one worked, just have to work out how to stop it posting to the database now if its already in there

kratsg · September 7, 2009

thank you, that last one worked, just have to work out how to stop it posting to the database now if its already in there

Change to:

if(mysql_num_rows($turn = mysql_query("SELECT letable WHERE id =$id LIMIT 1"))){
       echo 'that number exists already';
} else {
	if($id){
		mysql_query("INSERT INTO letable (`id`,`name`,`address`,`email`)VALUES('$id','$name','$address','$email')");
	}else{
	echo "<strong>id has not been filled in</strong>";
	}
}

saltedm8 · September 7, 2009

nah

im getting

Warning: mysql_num_rows() expects parameter 1 to be resource, boolean given in G:\wamp\www\lessons\homeworkdynamicform.php on line 32

again, and its still inserting

kratsg · September 7, 2009

Fix the query again xD I copied it wrong :-o

the "SELECT letable" to "SELECT * FROM letable"

saltedm8 · September 7, 2009

AHHHH.. 2 things.. its not putting anything into the database when its ok, and the second thing is when I test putting in no id

Warning: mysql_num_rows() expects parameter 1 to be resource, boolean given in G:\wamp\www\lessons\homeworkdynamicform.php on line 33

id has not been filled in

kratsg · September 7, 2009

Make sure it's this:

if(mysql_num_rows($turn = mysql_query("SELECT * FROM letable WHERE id =$id LIMIT 1"))){
	echo 'that number exists already';
} else {
	if(isset($id) && !empty($id)){
		mysql_query("INSERT INTO letable (`id`,`name`,`address`,`email`)VALUES('$id','$name','$address','$email')");
	}else{
		echo "<strong>id has not been filled in</strong>";
	}
}

saltedm8 · September 7, 2009

parse error line 98

kratsg · September 7, 2009

Copy/paste the whole code and put it in [php][/php] tags.

saltedm8 · September 7, 2009

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
<style type="text/css">
<!--
p{ margin:0; padding:0;}
form { margin-top:55px; }
input { margin: 4px 0 4px 0;}

.one {margin: 0 0 0 58px;}
.two {margin: 0 0 0 37px;}
.three {margin: 0 0 0 22px;}
.four {margin: 0 0 0 38px;}
.five {margin: 0 0 0 165px;}
-->
</style>
</head>

<?php 
include("connect.php");

$id =    $_POST['id'];
$name	= $_POST['name'];
$address	= $_POST['address'];
$email	= $_POST['email'];


if($_POST['submit']){

  
     if(mysql_num_rows($turn = mysql_query("SELECT * FROM letable WHERE id =$id LIMIT 1"))){

echo 'that number exists already';

} else {

if(isset($id) && !empty($id)){

mysql_query("INSERT INTO letable (`id`,`name`,`address`,`email`)VALUES('$id','$name','$address','$email')");	

}else{

echo "<strong>id has not been filled in</strong>";


}	

}
		  
		  



?>

<body>

<table width="50%" border="0">
  <tr>
    <td> </td>
    <td> </td>
    <td> </td>
    <td> </td>
  </tr>
</table>


<form id="form1" class="top" method="post" action="">
  <p>
    <label class="one">id:
      <input type="text" name="id" id="id" />
    </label>
  </p>
  <p>
    <label class="two">name:
      <input type="text" name="name" id="name" />
    </label>
  </p>
  <p>
    <label class="three">address:
      <input type="text" name="address" id="address" />
    </label>
  </p>
  <p>
    <label class="four">email:
      <input type="text" name="email" id="email" />
    </label>
  </p>
  <p>
    <label>
      <input class="five" type="submit" name="submit" id="submit" value="submit" />
    </label>
  </p>
</form>
</body>
</html>

bundyxc · September 7, 2009

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
<style type="text/css">
<!--
p{ margin:0; padding:0;}
form { margin-top:55px; }
input { margin: 4px 0 4px 0;}

.one {margin: 0 0 0 58px;}
.two {margin: 0 0 0 37px;}
.three {margin: 0 0 0 22px;}
.four {margin: 0 0 0 38px;}
.five {margin: 0 0 0 165px;}
-->
</style>
</head>

<?php 
include("connect.php");
$id = $_POST['id'];
$name = $_POST['name'];
$address = $_POST['address'];
$email = $_POST['email'];

if($_POST['submit']) {
if(mysql_num_rows($turn = mysql_query("SELECT * FROM letable WHERE id =$id LIMIT 1"))) {
	echo 'that number exists already';
} else {
	if(isset($id) && !empty($id)) {
		mysql_query("INSERT INTO letable (`id`,`name`,`address`,`email`)VALUES('" . $id . "','" . $name . "','" . $address . "','" . $email . "')");
	} else {
		echo "<strong>id has not been filled in</strong>";
	}
}
}
?>

<body>

<table width="50%" border="0">
  <tr>
    <td> </td>
    <td> </td>
    <td> </td>
    <td> </td>
  </tr>
</table>


<form id="form1" class="top" method="post" action="">
  <p>
    <label class="one">id:
      <input type="text" name="id" id="id" />
    </label>
  </p>
  <p>
    <label class="two">name:
      <input type="text" name="name" id="name" />
    </label>
  </p>
  <p>
    <label class="three">address:
      <input type="text" name="address" id="address" />
    </label>
  </p>
  <p>
    <label class="four">email:
      <input type="text" name="email" id="email" />
    </label>
  </p>
  <p>
    <label>
      <input class="five" type="submit" name="submit" id="submit" value="submit" />
    </label>
  </p>
</form>
</body>
</html>

saltedm8 · September 7, 2009

very close.... just got a warning when I tested not putting in any id at all

Warning: mysql_num_rows() expects parameter 1 to be resource, boolean given in G:\wamp\www\lessons\homeworkdynamicform.php on line 32
id has not been filled in

bundyxc · September 7, 2009

Try giving this one a shot.

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
<style type="text/css">
<!--
p{ margin:0; padding:0;}
form { margin-top:55px; }
input { margin: 4px 0 4px 0;}

.one {margin: 0 0 0 58px;}
.two {margin: 0 0 0 37px;}
.three {margin: 0 0 0 22px;}
.four {margin: 0 0 0 38px;}
.five {margin: 0 0 0 165px;}
-->
</style>
</head>

<?php 
include("connect.php");

if 
isset($_POST['id']) && 
isset($_POST['name']) && 
isset($_POST['address']) && 
isset($_POST['email']) && 
isset($_POST['submit']) && 
!empty($_POST['id'])) && 
!empty($_POST['name'])) && 
!empty($_POST['address'])) && 
!empty($_POST['email']))
{
if(mysql_num_rows(mysql_query("SELECT * FROM `letable` WHERE `id`='" . $_POST['id'] . "'LIMIT 1"))) {
	print('<b>Error:</b> Row of data with ID ' . $_POST['id'] . ' already exists in database.');
} else {
	mysql_query("INSERT INTO letable (`id`,`name`,`address`,`$_POST['email']`)VALUES('" . $_POST['id'] . "','" . $_POST['name'] . "','" . $_POST['address'] . "','" . $$_POST['email'] . "')");
}
} else {
print('<b>Error:</b> Form was not filled completely. Please try again.')
?>

<body>

<table width="50%" border="0">
  <tr>
    <td> </td>
    <td> </td>
    <td> </td>
    <td> </td>
  </tr>
</table>


<form id="form1" class="top" method="post" action="">
  <p>
    <label class="one">id:
      <input type="text" name="id" id="id" />
    </label>
  </p>
  <p>
    <label class="two">name:
      <input type="text" name="name" id="name" />
    </label>
  </p>
  <p>
    <label class="three">address:
      <input type="text" name="address" id="address" />
    </label>
  </p>
  <p>
    <label class="four">$_POST['email']:
      <input type="text" name="$_POST['email']" id="$_POST['email']" />
    </label>
  </p>
  <p>
    <label>
      <input class="five" type="submit" name="submit" id="submit" value="submit" />
    </label>
  </p>
</form>
</body>
</html>

saltedm8 · September 7, 2009

Parse error on line 34

lol.. we will get there hopefully

bundyxc · September 7, 2009

Alright, I ran it through my debugger, and it came through error-free. Let's just hope everything is clean in connect.php and your database so I can call it a night...

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
<style type="text/css">
<!--
p{ margin:0; padding:0;}
form { margin-top:55px; }
input { margin: 4px 0 4px 0;}

.one {margin: 0 0 0 58px;}
.two {margin: 0 0 0 37px;}
.three {margin: 0 0 0 22px;}
.four {margin: 0 0 0 38px;}
.five {margin: 0 0 0 165px;}
-->
</style>
</head>

<?php 
error_reporting(E_ALL);
ini_set('display_errors','1');

include("connect.php");

if (isset($_POST['id']) && isset($_POST['name']) && isset($_POST['address']) && isset($_POST['email']) && isset($_POST['submit']) && !empty($_POST['id']) && !empty($_POST['name']) && !empty($_POST['address']) && !empty($_POST['email'])) {
    if(mysql_num_rows(mysql_query("SELECT * FROM `letable` WHERE `id`='" . $_POST['id'] . "'LIMIT 1"))) {
        print('<b>Error:</b> Row of data with ID ' . $_POST['id'] . ' already exists in database.');
    } else {
        mysql_query("INSERT INTO `letable` (`id`,`name`,`address`,`email`) VALUES('" . $_POST['id'] . "','" . $_POST['name'] . "','" . $_POST['address'] . "','" . $$_POST['email'] . "')");
    }
} else {
    print('<b>Error:</b> Form was not filled completely. Please try again.');
}
?>

<body>

<table width="50%" border="0">
  <tr>
    <td> </td>
    <td> </td>
    <td> </td>
    <td> </td>
  </tr>
</table>


<form id="form1" class="top" method="post" action="">
  <p>
    <label class="one">id:
      <input type="text" name="id" id="id" />
    </label>
  </p>
  <p>
    <label class="two">name:
      <input type="text" name="name" id="name" />
    </label>
  </p>
  <p>
    <label class="three">address:
      <input type="text" name="address" id="address" />
    </label>
  </p>
  <p>
    <label class="four">$_POST['email']:
      <input type="text" name="$_POST['email']" id="$_POST['email']" />
    </label>
  </p>
  <p>
    <label>
      <input class="five" type="submit" name="submit" id="submit" value="submit" />
    </label>
  </p>
</form>
</body>
</html>

saltedm8 · September 7, 2009

very very close.. a little snag

Error: Form was not filled completely. Please try again.

is there before I even fill out the form, once its filled out its fine

bundyxc · September 7, 2009

Alright, here we go...

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
<style type="text/css">
<!--
p{ margin:0; padding:0;}
form { margin-top:55px; }
input { margin: 4px 0 4px 0;}

.one {margin: 0 0 0 58px;}
.two {margin: 0 0 0 37px;}
.three {margin: 0 0 0 22px;}
.four {margin: 0 0 0 38px;}
.five {margin: 0 0 0 165px;}
-->
</style>
</head>

<?php 
error_reporting(E_ALL);
ini_set('display_errors','1');

include("connect.php");

if (isset($_POST['id'])) && (isset($_POST['name'])) && (isset($_POST['address'])) && (isset($_POST['email'])) && (isset($_POST['submit'])) && (!empty($_POST['id'])) && (!empty($_POST['name'])) && (!empty($_POST['address'])) && (!empty($_POST['email']))
{
if(mysql_num_rows(mysql_query("SELECT * FROM `letable` WHERE `id`='" . $_POST['id'] . "'LIMIT 1"))) {
	print('<b>Error:</b> Row of data with ID ' . $_POST['id'] . ' already exists in database.');
} else {
	mysql_query("INSERT INTO letable (`id`,`name`,`address`,`$_POST['email']`)VALUES('" . $_POST['id'] . "','" . $_POST['name'] . "','" . $_POST['address'] . "','" . $$_POST['email'] . "')");
}
} elseif (isset($_POST['submit'])) {
print('<b>Error:</b> Form was not filled completely. Please try again.')
}
?>

<body>

<table width="50%" border="0">
  <tr>
    <td> </td>
    <td> </td>
    <td> </td>
    <td> </td>
  </tr>
</table>


<form id="form1" class="top" method="post" action="">
  <p>
    <label class="one">id:
      <input type="text" name="id" id="id" />
    </label>
  </p>
  <p>
    <label class="two">name:
      <input type="text" name="name" id="name" />
    </label>
  </p>
  <p>
    <label class="three">address:
      <input type="text" name="address" id="address" />
    </label>
  </p>
  <p>
    <label class="four">$_POST['email']:
      <input type="text" name="$_POST['email']" id="$_POST['email']" />
    </label>
  </p>
  <p>
    <label>
      <input class="five" type="submit" name="submit" id="submit" value="submit" />
    </label>
  </p>
</form>
</body>
</html>

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