listing help


Posted on 16th Feb 2014 07:03 pm by admin

Hi,

at first, great new design!!! I like it,

Did you know?Explore Trending and Topic pages for more stories like this.
i got a problem,

I want to list some tekst, but there is a users, users can list they own texts, and all teksts, and find words in all texts and in own texts

somethings is weong in my php code because ist dont work

At first here is the form file:

Code: <div>
<form name="TextSearchForm" action="<?php print "list_txt_search.php";?>" method="post" target="main">

<div><select name="sn_text_year" style="text-align: center;">
<option selected="selected"> Year </option>
<option value="2009">2009</option>
<option value="2010">2010</option>
<option value="2011">2011</option>
</select>
</div>

<div><select name="sn_text_author" style="text-align: center;">
<option selected="selected"> choose </option>
<option value=""> all txt </option>
<option value="<?php print $UsrNrID;?>"> my txt </option>
</select>
</div>

<div><input type="text" name="TextSearch" size="30" /></div>
<div><input type="submit" name="SubmitTxtSerch" value="LIST" class="Button" /></div>

</form>
</div>

3 selection is there:
"sn_text_year" (which year is text wroten)
"sn_text_author" (who is the author of the text - this is a cookie whith name of the user)
"TextSearch" (if the user want to find some word in database)


Code: <?php
$TextSearch = $_POST['TextSearch'];
?>



<?php
// UsrIdentify
$query_user = mysql_query("SELECT * FROM `sn_users` WHERE `sn_users_username` = '".$_COOKIE['loggedin']."'");

if(!$query_user){
print mysql_error();
exit;
}

$RequestUsrID = mysql_fetch_array($query_user);


$UsrID = $RequestUsrID['sn_users_id']; // HERE I CAN TAKE THE USERS ID
?>



<?php

if ( $TextSearch == '' && $UsrID ) {
$QeryUsr = "SELECT * FROM `sn_text` WHERE `sn_text_godina` = '" . $_POST['sn_text_year'] . "' AND `sn_text_author` = '" . $UsrID . "' ORDER BY `sn_text_id` DESC";
}

if ( $TextSearch == '' && $UsrID == '' ) {
$QeryUsr = "SELECT * FROM `sn_text` WHERE `sn_text_year` = '".$_POST['sn_text_year']."' ORDER BY `sn_text_id` DESC";
}

if ( $TextSearch && $UsrID == '' ) {
$QeryUsr = "SELECT * FROM `sn_text` WHERE ucase(`area1`) LIKE '%".$TextSearch."%' OR lcase(`area1`) LIKE '%".$TextSearch."%' AND `sn_text_year` = '".$_POST['sn_text_year']."' ORDER BY `sn_text_id` DESC";
}

if ( $TextSearch && $UsrID ) {
$QeryUsr = "SELECT * FROM `sn_text` WHERE ucase(`area1`) LIKE '%".$TextSearch."%' OR lcase(`area1`) LIKE '%".$TextSearch."%' AND `sn_text_year` = '".$_POST['sn_text_year']."' AND `sn_text_author` = '" . $UsrID . "' ORDER BY `sn_text_id` DESC";




// Admin capacityes
$query_user = mysql_query( $QeryUsr );

if(!$query_user){
print mysql_error();
exit;
}

while($request = mysql_fetch_array($query_user)) {

// list
}
?>


As You see You can choose:

1. Only year of text wroten
2. No text search, but identify User
3. No text search, no User identification - mean ilst all text
4. Search text, and identify User (mean searching word only in Users texts)

So, something is wrong, because:
1. when I choose Year, and list all text - list only my texts
2 when I choose Year, and list my texts - list only my texts
3. when I choose Year, and list my texts, with searching text - list only my text without choose
4. when I choos Year, and list all text, with serching text - its work fine

I see its a complicated, I hope, that You can give mi a solution

thanx
No comments posted yet

Your Answer:

Login to answer
320 Like 27 Dislike
Previous forums Next forums
Other forums

Problems with returning true or false in eval()'d code
Hi guys,

Would appreciate some help with a problem when running eval() on a function that sho

Add User script "Could not execute query"
This should be an easy script but I can't get it to run. Can someone please help me?

<

Email Processor
I have a few questions so this post will be a larger one! Sorry, but I'm a bit of a PHP newbie so be

square instead of number
Hello
I do not know why but this code seems to work fine only in my xampp local insallation but n

Alternate messaging
I have 4 strings in MySQL db1

$string1 : Hello
$string2 : Hi
$string3 : Great
$strin

Create PHP table grid help, please
I have a MySQL database setup, now it's time for the table (gridview) design in php. Here's the prob

split values
I have values that are returned to me in this format:

name=>test,age=>49

code help - pagination
Hi all, I have this code, basically a user logs into my site and they get this page.

The pro

couldn't connect to your database
Hello I am new to php mysql

Actually i have read A tutorial on nettuts
"http://net.tu

PHP - MySQL Fail
My PHP code will only execute the first part of my code...

Code: <?php

sessio

Sign up to write
Sign up now if you have flare of writing..
Login   |   Register
Follow Us
Indyaspeak @ Facebook Indyaspeak @ Twitter Indyaspeak @ Pinterest RSS



Play Free Quiz and Win Cash