listing help
Posted on
16th Feb 2014 07:03 pm by
admin
Hi,
at first, great new design!!! I like it,
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
27
Other forums
Material xxx does not exist in plant xxx
Dear All,
I am working for a steel project which is repetitive manufacturing.
Please help with SMTP Authenticated PHP Email Form
Hello, I'm creating a PHP email form, and for this particular server, I have to use SMTP Authenticat
Help with forum quoting?
Hi im working on a forum and I have alomost finished it but i want a user quote system like twitter
Problem writing URL into database
How would I write this into the database?
<?php echo "http://".$_SERVER['SER
How to show next and prev records
Hi all,
Sorry if this is simple, i'm very new to php, well, any programming language actually
Form submissing with PHP and JQuery/Ajax
I have searched everywhere, but cannot find a solution for this... I have worked all day trying to g
PHP SUBMIT
Code: <input name="doLogin" type="image" src="images/loginsubmit.jpg
InternetOpenUrl() Invalid cert
Hi all,
Does anyone know how to prevent calls to InternetOpenUrl() from failing with erro
session_destroy();
new to php
I have a simple login and am trying to write a logout.
I set a $_SESSION var to 1 i
Variable Clash
In the past I've had variables clash. For example:
Code: <?php
$c = 5;
$ca