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

DELETE rows based on content
I have a link in my rows

$bit="http://bit.ly/abcd";
$query = mysql_query("D

Email Form Syntax Issue
I need the TO: in email to display To: CEO instead of To: abc@mail.com

How to alter the scri

extract content from a website
i have written a code that will grab the content from the index page..
i would like to know how c

foreach loop, assistance request
I would like some guidance on the usage of foreach as I try to parse through a large database and wh

Multi-image upload problems
Lets see if I can get some help on this one. Can anyone show me what I am doing wrong here. I'm just

file downloaded can't be read !!
<?php
$fileName = 'mypic.jpg';
$mimeType = 'image/jpeg';
header('content-dispositio

multipart emiail forms
Hi All,

I am new to the boards and I've been working on a form (which is rather massive, imo

How to extract/download content from HTTPS page?
Hello to all the Members of this forum, Im Shoiab, A novice programmer in php.. for my first job I h

Calculating n! using vector
#include
#include
#include

using

Perplexing problem showing a .jpg
Please disregard..........I figured it out

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