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General PHP Help
#1

Dear Sir I have following codes

[php]<?php
require_once(“connect.php”);

if(isset($_POST[‘button1’]))
{
// Get values from form
$sno =$_POST[‘txtsno’];

	$record_check ="SELECT * FROM test WHERE sno = " . $sno;
	$result=mysqli_query($record_check);
	$row = mysqli_fetch_array($result) ;

	if(!$row)
				die ('No record Found'); 
}

?>

Display data in textboxes

html {
overflow:auto;
}

body {
background-color:#FFFFFF;
margin:0 auto;
}

#mypopup
{
float: left;
width: 250px; height: 350px;
background: #90bade;
border: 1px solid #069;
text-align:center;
padding:2px;
margin-top:150px;
margin-left:100px;
overflow:auto;
}

#header
{
background-color:#3399FF;
background-position:left center;
line-height:25px;
font-size:22px;
color:#FFFF33;
font-weight:600;
border-bottom:1px solid #6699CC;
padding:10px;
}

Search Data
 <form name="form1" action="data_find_in_textbox2.php" method="post">
    <table border=0; cellpadding="1" cellspacing="1" bgcolor="#CCFFFF" align="center" >
      <tr>
        <td>Code</td>
        <td width="50px"><input type="text" name="txtsno" value="" title="Enter product code" onkeypress="validate(event)" ;  onfocus="this.select()" /></td>
      </tr>
      <tr>
        <td>Product</td>
        <td><input type="text" name="txtpro" value="<? echo $row['packing'] ?>" title="Enter product name" ></td>
      </tr>
      <tr>
        <td>Weight</td>
        <td><input type="text" name="txtwet" value="<? echo $row['weight'] ?>" title="Enter product weight" onfocus="this.select()" ></td>
      </tr>
    </table>
    <div style=text-align:center;margin-top:20px;>
      <input type="submit" name="button1" value="Display" >
      <input type="submit" name="button2" value="Clear" >
    </div>
  </form>
</div>
<?php mysqli_close($con); ?> [/php]

I want to search data like shown in attachment.

If I enter sno and press Dispaly button, then text2 and text3 must display relevant data from table.

Now when enter sno=1 and press Display button then it shows this error message

Warning: mysqli_query() expects at least 2 parameters, 1 given in C:\wamp\www\db\data_find_in_textbox2.php on line 10

Test table has more than 3 recores including sno=1

Please help

search.jpg

#2

Look up the mysqli_query documentation, or search for that error message.

http://us1.php.net/mysqli_query

Object oriented style mixed mysqli::query ( string $query [, int $resultmode = MYSQLI_STORE_RESULT ] )

Procedural style
mixed mysqli_query ( mysqli $link , string $query [, int $resultmode = MYSQLI_STORE_RESULT ] )

since you are using procedural style (I would recommend changing to object oriented) you should send in two parameters, first the mysqli connection, then the query.

[php]$result=mysqli_query($conn, $record_check);[/php]

[hr]

Your code is vulnerable to sql injection which means anyone and everyone can read/dump your entire database. You should change to use parameterized queries immidiatly.

#3

Thanks Sir for good tips but…

the problem still not solved.
Please make some more modifications in my code.

Thanks again

#4

The error message was for only sending in one parameter so you couldn’t possibly still get the same error message. Please post the error message you get along with relevant code (connect.php and the same file as you posted last time). Remember to remove login credentials before posting.

#5

Sir now i have following modified codes, but when i press Display button then it displays no error but also does not show any data in other textboxes relevant to first textbox

[php] <?php
global $con;
require_once(“connect.php”);

if(isset($_POST[‘button1’]))
{
// Get values from form
$sno =$_POST[‘txtsno’];

	$record_check ="SELECT * FROM test WHERE sno = " . $sno;
	$result=mysqli_query($con,$record_check);
	$row = mysqli_fetch_array($result) ;

	if(!$row)
				die ('No record Found'); 
}

?>

Search Data

html {
overflow:auto;
}

body {
background-color:#FFFFFF;
margin:0 auto;
}

#mypopup
{
float: left;
width: 250px; height: 350px;
background: #90bade;
border: 1px solid #069;
text-align:center;
padding:2px;
margin-top:150px;
margin-left:100px;
overflow:auto;
}

#header
{
background-color:#3399FF;
background-position:left center;
line-height:25px;
font-size:22px;
color:#FFFF33;
font-weight:600;
border-bottom:1px solid #6699CC;
padding:10px;
}

Search Data
  <form name="form1" action="data_find_in_textbox_Jiml.php" method="post">
     <table border=0; cellpadding="1" cellspacing="1" bgcolor="#CCFFFF" align="center" >
       <tr>
         <td>Code</td>
         <td width="50px"><input type="text" name="txtsno" value="" title="Enter product code" onkeypress="validate(event)" ;  onfocus="this.select()" /></td>
       </tr>
       <tr>
         <td>Product</td>
         <td><input type="text" name="txtpro" value="<? echo $row['packing'] ?>" title="Enter product name" ></td>
       </tr>
       <tr>
         <td>Weight</td>
         <td><input type="text" name="txtwet" value="<? echo $row['weight'] ?>" title="Enter product weight" onfocus="this.select()" ></td>
       </tr>
     </table>
     <div style=text-align:center;margin-top:20px;>
       <input type="submit" name="button1" value="Display" >
       <input type="submit" name="button2" value="Clear" >
     </div>
   </form>
 </div>
<?php mysqli_close($con); ?>

[/php]

search.jpg

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