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

Insert Failing.
Hey,
I am making a Sign up page for a website, but the insert query into the Database does not se

Using system() and bringing back the results
I am aware that you can use system() within PHP to execute system commands, but I was wondering if t

Display search result
Hi!

I have a SQL database with information about albums and track (music).

This is wh

UL and LI Add Form
The idea I want here is when the user click on a character name from the drop down select bar at the

$_POST variable un-useable
I'm trying to use a $_POST variable in a mysql update statement but i can't use it for some unknown

Problem with the Update command used with a sqldataadapter
I'm connected to a database on an SQL Server and I'm using a sqldataadapter, sqlconnection, sqldatas

Rounding a number queried from a database
I know that to display a rounded number you just do echo "round($number)";. But how would

Mass activity scheduling
Hi experts,

Im facing some problems to deal with mass activity jobs.
When I run some

What do you call the "token" thing?
You know how some sites have links that run on tokens? Tokens are links that only stay alive for a c

This is driving me nuts!
This insert query looks to be alright, however I get this error:

QuoteYou have an error in yo

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