a listbox populated with query strings when selected particular query to be executed

27 New Member

hi all,

somebody help me out on this

i have a listbox where i have the values as starts with and ends with and next to that is a textbox where we give the input whether the fieldname should start with a particular letter or end with a particular letter

say for example the field name lastname is to be executed

so what we do is select a value from listbox say startswith and in the textbox we give as letter p and that should execute a mysql query that the lastname should start with letter p and display all the lastnames which all starts with letter p

hope could be understood

thanks in advance

Apr 24 '09 #1

Subscribe Reply

1764

Dormilich

8,658

Recognized Expert Moderator Expert

this sounds like a hell of validation to do… anyways you might want to look here on string comparison in MySQL.

Apr 24 '09 #2

Ciary

247

Recognized Expert New Member

i think you better write 3 SQL Queries looking like this:

Expand|Select|Wrap|Line Numbers

 
if($_POST['validateType'] == 'start'){

   $query = "SELECT * FROM yourtable WHERE lastname LIKE " . $_POST['myval'] . "%";

} if($_POST['validateType'] == 'end'){

   $query = "SELECT * FROM yourtable WHERE lastname LIKE %" . $_POST['myval'];

} else {

   $query = "SELECT * FROM yourtable WHERE lastname LIKE %" . $_POST['myval'] . "%";

}

i'm not sure if this is what you need but it generates a recordset depending on what you selected

Apr 24 '09 #3

ahilar12

New Member

hi dormi,

could u clarify me what these variables represent in this following code

1.validateType
2.start
3.myval

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      if($_POST['validateType'] == 'start')

          {

    $query = "SELECT * FROM yourtable WHERE lastname LIKE " . $_POST['myval'] . "%";

 } if($_POST['validateType'] == 'end'){

    $query = "SELECT * FROM yourtable WHERE lastname LIKE %" . $_POST['myval'];

 } else {

    $query = "SELECT * FROM yourtable WHERE lastname LIKE %" . $_POST['myval'] . "%";

 }

Apr 27 '09 #4

Dormilich

8,658

Recognized Expert Moderator Expert

@ahilar12

where to have the search sting ("start": at the beginning, "end": at the end, anywhere in the text)
see above
the search term

Apr 27 '09 #5

ahilar12

New Member

hi ciary
check out this code and tell me whats wrong

Expand|Select|Wrap|Line Numbers

 
<html>
 
<head>

<script language="javascript">

function ShowHideForm(o){

    if(o.checked == true)

        document.getElementById("formId").style.display="block";

    else

        document.getElementById("formId").style.display="none";

}

</script>

<body>
 
<?php

include 'menu.php';

?>
 
<form name="createreport" action="insertreport.php" method="POST">

<input type="checkbox" name="test" onclick="ShowHideForm(this)"> First Name

<br>

<br>

<div id="formId" style="display:none">

<table border="1">

  <tbody>

    <tr>

      <td>FirstName</td>

      <td><input type="text" name="myval"/>   </td>

      <td><select name="validatetype">

      <option name="start">start</option>

      <option name="end">end</option>
 
</select>

</td>
 
<td><input type="submit" value="submit"/></td>
 
    </tr>

  </tbody>

</table>
 
</div>

</form>
 
</body>

</html>

Expand|Select|Wrap|Line Numbers

  
<?php
 
       include 'connect.php';
 
       $ins=mysql_query("insert into createreport(validatetype,myval)values('$_POST[validatetype]','$_POST[myval]')");
 
if($ins)

{

  echo "inserted";

}
 
if(!$ins)

{

    echo "mysql_error()";

}
 
      if($_POST['validatetype'] == 'start')

  {

    $query = mysql_query("SELECT * FROM createreport WHERE lastname LIKE " . $_POST['myval'] . "%");

    echo $query;

 } 

 else if($_POST['validatetype'] == 'end')

{

    $query = mysql_query("SELECT * FROM createreport WHERE lastname LIKE %" . $_POST['myval']);

    echo $query;

 } else 

{

    $query = mysql_query("SELECT * FROM createreport WHERE lastname LIKE %" . $_POST['myval'] . "%");

    echo $query;

 }

?>

Apr 27 '09 #6

Dormilich

8,658

Recognized Expert Moderator Expert

@ahilar12
there's no input validation (see SQL Injection). and some indices are not quoted.

what is the code not doing / doing, what it should do / not do?

are there any error messages?

Apr 27 '09 #7

Ciary

247

Recognized Expert New Member

what does it say as error? or what do you see?

ps. try codeblocks, makes it easier to read

Apr 27 '09 #8

ahilar12

New Member

hi ciary,

what is code blocks?

i m a beginner to php.

Apr 27 '09 #9

Dormilich

8,658

Recognized Expert Moderator Expert

codeblocks belong to the forum to make source code easier to read.

Expand|Select|Wrap|Line Numbers

// this is a code block

Apr 27 '09 #10

Ciary

247

Recognized Expert New Member

it's not php. it's when you post code on the forum. on top of your editor when replying, you see a symbol like this: #. it will add codeblocks. then put your code between them.

Apr 27 '09 #11

Markus

6,050

Recognized Expert Expert

[code] code goes here [/code].

[PHP] and [HTML] used to do the same thing, but no more. Please use the above.

Moderator.

Apr 27 '09 #12

ahilar12

New Member

oh ok

understood

thanks

Apr 27 '09 #13

Ciary

247

Recognized Expert New Member

np :)

now, you said something about an error in the code. what does it do? is that what you want it to do? are there any errors?

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a listbox populated with query strings when selected particular query to be executed

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