[[SOLVED] How to SELECT item in lstbox?]

use

if ($selectedItem == 'value') { echo ' selected="selected" '; }

 

put that line of code in each <option> tag and change the value for each option

[[SOLVED] Issue with PHP file]

try echoing $_POST['EnglishWord'] and $_POST['EnglishDefinition']..

and also echo the query to check its all correct

i.e.

<?php
   /** PHP code inserted here */
   $greekWord = $_POST['GreekWord'];
   $greekDef = $_POST['GreekDefinition'];
   $englishWord = $_POST['EnglishWord'];
   $englishDef = $_POST['EnglishDefinition'];
   $chapter = $_POST['chapter'];
   $section = $_POST['section'];

//test to see if values being posted...
echo $_POST['EnglishWord'];
echo '<br />';
echo $_POST['EnglishDefinition'];
            
   echo "connection to the DB section of the code";
   $conn = @mysql_connect("localhost", "root") or die("Could not connect to the data base");
   $db = @mysql_select_db("test",$conn) or die("Error in selecting the data base.");
   
   echo "setting the SQL variables up";
   // This is so the SQL statement is correctly formatted.
   $grkWrd = "'".$greekWord."'";
   $grkDef = "'".$greekDef."'";
   $engWrd = "'".$englishWord."'";
   $engDef = "'".$englishDef."'";
   $chptr = "'".$chapter."'";
   $sctn = "'".$section."'";
   
   echo "the query is here";
   $query = "INSERT INTO word (GreekWord, GreekDefinition, EnglishWord, EnglishDefinition, Chapter, Sections) values ($grkWrd,$grkDef,$engWrd,$engDef,$chptr,$sctn)";

// echo query to validate
echo '<br />';
echo $query;
   
   echo "exuctuing the query";
   $exec = @mysql_query($query, $conn) or die ("Cannot add the word to the database");         
   
   echo "query execution successful";
   if($exec)
   {
      //echo "<a href=\"javascript:history.back();\">Please go back and fill out the missing fields</a>";
      echo "Word added";
      //header("Location:http://localhost/ninjatuna/wordAdded.html");
   }
   else
   {
      if($errmsg!="")
      {
         echo($errmsg);
         echo "<a href=\"javascript:history.back();\">Please go back and fill out the missing fields</a>";
         exit;
      }
   }
?>

[date/mysql question]

maybe have a seperate column creationDate? and then you can determine if it was created before that quarter or if they just didnt' visit...

another option would be to make the previous quarters NULL if a store is created after the first quarter..

i.e. if its made in the 3rd quarter, first and second quarter values would be NULL and the third would have a date..?

i think my first suggestion is the most practical tho

 

edit: rhodesa beat me to it! wrote too much to delete the post tho...

[[SOLVED] A converting a string backwards?]

you can split the string up into an array then reverse each element of the array then implode back into a string...

 

$string = "this is a test string!";

$array = str_split($string, 2);

foreach ($array as $key => $value) {
$array[$key] = strrev($value);
}

//you can leave as an array or implode back into a string
$string = implode($array);

[Match end of string]

explode[2] would work alrite, or you could use some of the string functions and get rid of "kick user" leaving you the id...

 

what are you trying to do here?

 

might be an idea to store it in an array.. i.e.

$test[1] = "kick user";

$test[2] = "userID";

 

...?

[[SOLVED] PHP - After file upload, send user e-mail]

first up you have two form elements called 'email'... and you need to pass the URL on to the email script, so you need a hidden element with the url

<input type="email" name="email" id="email" size="40"><p align="center"><label for="email">  <input type="submit" name="email" />

 

change to

<input type="email" name="email" id="email" size="40">

<p align="center"><input type="submit" name="submit" />

<input type="hidden" name="url" value="'. $url2 . $url . $filename . '" />

 

 

make the form action go to say

sendmail.php

which would look like

<?php
if (isset($_POST['email'], $_POST['url']) && !empty($_POST['email']) && !empty($_POST['url'])) {
$to = $_POST['email'];
$url = $_POST['url'];
$subject = 'File Upload URL';
$message = "Your URL is: $url";
$from = "youremail@yourdomain.com";
$headers = 'From: $from' . "\r\n" .
    'Reply-To: $from' . "\r\n" .
    'X-Mailer: PHP/' . phpversion();

mail($to, $subject, $message, $headers);
}

//redirect user to whatever page
header("location: index.php");
exit();
?>

 

obviously change the message / subject / from etc to whatever you want... and also make sure you change the index.php in the header/redirect to whatever page you want them to be forwarded to

 

[[SOLVED] using a varialbe inside a php email script with html]

you can either change the ' to ", so that php parses all the variables within the quotations...

or wherever the variable goes you can put it in like '.$variable.'

 

i.e. 1

$message = "This is the email script: $variable etc etc";

 

i.e. 2

$message = 'This is the email script: '.$variable.' etc etc';

[mail with Attachements not working]

@the eagle

you wrote this?

 

@seventheyejosh

this is the same/similar to a mail script I have been using... have you had any troubles with it and gmail?

[Noob help]

It's likely not the issue, but it is common to some scripts, and it may have been the issue, most likely not.

 

hahaha

[[SOLVED] form output]

your input code needs to be inside a form...

i.e.

<div class="wrapSearch">
  <div>
<form action="<?php echo basename($_SERVER['PHP_SELF']); ?>" method="post">
  <input class="input" name="input" type="text" id="input1" size="30" maxlength="1000" onkeyup="autoSuggest(this.id, 'listWrap1', 'searchList1', 'input1', event);" onkeydown="keyBoardNav(event, this.id);" />   
<input type="submit" name="search" id="search1" value="Search" />
</form>
  </div>
  <div class="listWrap" id="listWrap1">
  <ul class="searchList" id="searchList1">
  </ul>
  </div>
</div>

 

that form will post the contents to itself.. and the php will pick it up.. put the php in the very top of the page! before any <html> <head> etc

 

if (isset($_POST['input']) && !empty($_POST['input']) {
$userInput = $_POST['input'];
//change spaces to -
$userInput = str_replace(' ', '-', $userInput);

//redirect user to url
header("location: page/$userInput");
exit();
}

 

that will redirect the user to www.mydomain.com/whatever/page/user-input

what did you want to try and achieve from this?

[[SOLVED] Decrypt sha1 passwords?]

if you want the ability to decrypt/encrypt them, next time use mysql's AES_ENCRYPT() function.

http://dev.mysql.com/doc/refman/5.1/en/encryption-functions.html

[[SOLVED] form output]

if you had a form with text input called say userInputBox

 

you could have

 

$userInput = $_POST['userInputBox'];
//change spaces to -
$userInput = str_replace(' ', '-', $userInput);

//redirect user to url
header("location: page/$userInput");
exit();

[changing time by two hours]

use this code to change the timezone...

putenv('TZ=Australia/Sydney');

 

obviously change Australia/Sydney to whatever timezone

 

a list of supported timezones is here

http://au2.php.net/manual/en/timezones.php

[[SOLVED] Undefined Variable/Supplied Argument Is Not Valid]

you use

$get_items_res

and then

$get_item_res...

 

 

[[SOLVED] Undefined Index?]

does your url have www.whatever.com?cat_id=whatever in it?

 

you should probably also change it to

 

if(isset($_GET['cat_id']) && $_GET['cat_id'] == $cat_id) {

 

 

[How to sort vehicles without going to another page?]

iframe's would take almost as much time as re-loading the entire page, and will require unnecessary PHP/MySQL calls when you can do it all instantly with jquery/javascript....

 

try this jquery plugin... its pretty simple you should get the hang of it

 

http://tablesorter.com/docs/

[How to search a database, then have the option to sort by year, color etc?]

say you had a table called people:

id, name, suburb, postcode, age

 

and you were trying to find people in a suburb...?

your form would have

 

suburb: _______ [text input]

order by: ______ [drop down with options: name, postcode, age -- must correspond to the SQL columns]

 

then your php would be

$suburb = $_POST['suburb'];
$orderBy = $_POST['orderBy'];

$sql = @mysql_query("SELECT * FROM people WHERE suburb = '$suburb' ORDER BY $orderBy");

 

or if you wanted more leniancy with teh search you could use LIKE... i.e.

 

$sql = @mysql_query("SELECT * FROM people WHERE suburb LIKE '%$suburb%' ORDER BY $orderBy");

 

the %'s are wildcards...

[Invalid argument supplied for foreach()]

you should use $_GET / $_POST instead of $_REQUEST.

 

I don't know exactly how the script works, but I'm guessing not all of the $_REQUEST elements are arrays - hence the second foreach is causing an error..

 

try

 

if(is_array($_REQUEST)){
   foreach ($_REQUEST as $sel){
if (is_array($sel) {
      foreach($sel as $sel_product => $id)
      {
         echo "Selected product:" . $sel_product. "<br>";
         echo "Selected id:" . $id. "<br>";
      }
}
        }
}

[[SOLVED] javascript/jquery POST to recognize retrieved elements...]

got it working with the jquery 'live' function!

thanks dj kat & thorpe

[Creating monthly sales report]

how are the sales stored?

in a database? are the sales online?

[[SOLVED] Alert notification that user's got new message.]

use the window.onload function, so it only executes once the page is loaded.

 

<script type="text/javascript">
window.onload = function() { 
alert('You have unread comments.');
    }; 
</script>

[Ignore last 4 characters]

but if it were a file like 'bedroom.old.jpg', explode wouldn't work?

well it would, but you'd only get 'bedroom' and not 'bedroom.old'

you can use the pathinfo() function

http://us3.php.net/manual/en/function.pathinfo.php

 

$file = 'bedroom.jpg';

$filename = pathinfo($bedroom, PATHINFO_FILENAME);

[Database Table not Updating]

Error message = You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ' url = WHERE id =' at line 1

 

url = WHERE id =,

none of the variables are being sent in the url?

...

does your url look like

whatever.php?id=X&url=Z&linkname=Y

 

x, y and z are obviously the user input from the field.

 

you should change the db query to the following in case the variables arent sent.

//check if linkname, url and id are sent in the URL
if (isset($_GET['linkname'], $_GET['url'], $_GET['id']) {
$update = mysql_query("UPDATE nav SET linkname = " . $_GET['linkname'] . " , url = " . $_GET['url'] . " WHERE id = " . $_GET['id'] . ";");
}

 

 

 

 

[Submitting data via POST, but without the form]

use jquery.

its amazing!

jquery.com

[[SOLVED] javascript/jquery POST to recognize retrieved elements...]

I have a page which allows users to edit rows in the table,

When they click modify it uses a Jquery/javascript $.post function

This posts to a PHP page which sends back a form.

This all works fine, however, in the form I have a drop down menu where if certain values are selected I need different options to be shown/hidden - this works fine on elements that are on the page when it loads, but the jquery doesn't seem to recognize any of the elements in the retrieved form

 

Take for example the following:

 

HTML Page

 

<html>
<head>
<script type="text/javascript">
		jQuery(document).ready(function() {
		$('.editItem').click(function(event) {
				event.preventDefault();

				$.post("test.php", {}, 
					   function(data) {
						   if (data.length > 0) {
							   $('#testDiv').html(data);
						   } else {
							   $('#testDiv').html("<div align='center'>Error retrieving item details</div>");
						   }
					   });

		});

//test function for testButton
		$('.testButton').click(function(event) {
			event.preventDefault();
			alert('This test works!');
	  	});

});
</script>
</head>
<body>
<form id="test">
<button id="formRetrieval">Retrieve Form!</button>
<button class="testButton">Test Function!</button>
</form>
<div id="testDiv">
</div>
</body>
</html>

 

 

this is the php file the jquery posts to... test.php

<?php 
//just a dummy script to echo the form
echo '<form id="test">
<button class="testButton">Click me!</button>
</form>';
?>

 

 

In the above, if I've typed it all correctly, only the first TestFunction button will display an alert (the one on the HTML page).

When the 'Retrieve Form' button is clicked, it will load a new form into the testDiv and there will be a button with the same class 'testButton' (same as teh test function button), however, it will not display the alert.

 

If I had written a function, say, showAlert() which showed an alert, and set the button to onclick="showAlert()" then it would work.

 

In other words, the classes/id's of the elements which are being loaded through the PHP / post function aren't being identified...

 

Hope this all makes sense,

 

Any help would be greatly appreciated!