Deanznet

Okay. Lets say i have a list box. <select name="drop1" id="Select1" size="4" multiple="multiple"> <option value="1">item 1</option> <option value="2">item 2</option> <option value="3">item 3</option> <option value="4">item 4</option> <option value="0">All</option> </select> Now what i need it to do is open a file and replace the string [Replace] with whats in the list box. But it has to loop through everything in the list box understand? So How can i make it so that after it replaces 2 strings it gose to the next list item ect. after it replaces 2 more strings it goes to the next. So basically every 2 strings it will switch! Please help! here my code so far. <?php $fh = fopen(" myfile.txt", "r+"); if($fh==false) die("unable to create file"); ?> Please help!

This shows the Username and the Money they have in the account. Now heres my question.. How can i assort the Value row so i can have it in from highest to lowest? I tried order By but it did not work.. The data is in 2 different tables using a Join i think . HELPS! <? include ("../conf/conn.php"); ?> <? function current_balance($userid) { $query = "SELECT SUM(value) FROM user_banking WHERE userid='".$userid."'"; $result = mysql_query($query) or die(mysql_error()); // Print out result while($row = mysql_fetch_array($result)) { if($row['SUM(value)'] == 0) { return 0; } else { return $row['SUM(value)']; } } } ?> <?php $sql = mysql_query("SELECT * FROM users") or die(mysql_error()); while($data = mysql_fetch_array($sql)) { if(current_balance($data['id']) > 0) { ?> <table> <tr> <td><?php echo $data['login']; ?></td> <td><?php echo current_balance($data['id']); ?></td> </tr> <?php } } ?>

Thanks! Got it working with this tutorial here: w-w-w.w3schools.com/PHP/php_ajax_database.asp But i just cant figure out to add a second query.

First Question I have a database setup Id Item Cat1 Cat2 2 Shirt offroad gloves Something along those lines Now i want to be able to have 2 select boxes like this and when you click on the second one it will automatically search the Mysql records something that has offroad in cat1 and what ever you select in the other box for cat2 Help Please? Ha Here is my second question. a little different db setup Id Item Cat1 Cat2 cost retail 2 Shirt offroad gloves $1.99 Now i have php page i made to add this but i want to say lets if im putting the item in the glove category i would want to add 20% to the 1.99 and put that in the retail row but if i stock it in maybe the hat category i might want to add 40% understand? Any idea how?

Okay whats this part? and how do i decode this? <script> var _0x4b77=["\x73\x72\x63","\x6F\x66\x66\x65\x72\x66\x72\x61\x6D\x65","\x67\x65\x74\x45\x6C\x65\x6D\x65\x6E\x74\x42\x79\x49\x64"];function check_gm(){if(document[_0x4b77[2]](_0x4b77[1])[_0x4b77[0]]!=_0x4b77[3]);setTimeout(_0x4b77[6],1000);} ; </script> Can someone tell me how i can decode this? and how did they encode it in the first place? Thanks! If not can someone tell me how i can encode my js like this?!

Hey thanks! But i just got more problems lol! See if you can understand this. I need to post this information for to the page. <form method="post" action="http://site.com/Firststep1.php" onsubmit="return check();"> <input class="n" id="a" type="text" value="" size="3" maxlength="3" name="a"/> ) <input class="n" id ="p" type="text" value="" size="3" maxlength="3" name="p"/> <input class="n" id="n" type="text" value="" size="4" maxlength="4" name="n"/> <input type="submit" value="submit" name="submit1"> </form> before the results i want to grab are displayed. After that is posted to the page I want to grab the info on Firststep1.php <input type="hidden" name="animal" value="zebra"> <input type="hidden" name="color" value="white"> And echo it like this Zebra:White I think i need to use Curl any idea?!@!

Okay how do i screen scrap hidden input fields? <input type="hidden" name="animal" value="zebra"> <input type="hidden" name="color" value="white"> Can someone show me a code? Im not good with pregmatch ! thanks!

Hey thanks! but still cant get it to work.. Could the white page be a time out?? Im on godaddy shared hosting, If i use a GoDADDY proxy server like this curl_setopt ($ch, CURLOPT_PROXYTYPE, CURLPROXY_HTTP); curl_setopt ($ch, CURLOPT_PROXY,"http://64.202.165.130:3128"); It works. but i cant use any other proxies ?

Okay wait so i need to add curl_setopt($cUrl,CURLOPT_RETURNTRANSFER,1) ? Do i take out my $PageContent = curl_exec($cUrl); curl_close($cUrl); echo $PageContent; ??

I tried long tags <?php and such.. But still doesn't work. Is it the proxy's timing out or what? Maybe my host? Anyone get it to work on their servers?

the Proxy is a HTTP and it is good i check it? Just a white page?!@ <? $url = 'http://www.whatismyip.com/'; $cUrl = curl_init(); curl_setopt($cUrl,CURLOPT_VERBOSE,1); //curl_setopt($cUrl,CURLOPT_HTTPPROXYTUNNEL,true); curl_setopt($cUrl,CURLOPT_PROXYTYPE,CURLPROXY_HTTP); curl_setopt($cUrl,CURLOPT_PROXY,'68.32.66.6:9090'); curl_setopt($cUrl,CURLOPT_SSL_VERIFYPEER,false); curl_setopt($cUrl,CURLOPT_URL,$url); curl_setopt($cUrl,CURLOPT_TIMEOUT,320); $PageContent = curl_exec($cUrl); curl_close($cUrl); echo $PageContent; ?>

Because i cant use the same Proxy again thats why it has to be removed from the txt file.

Okay here is the thing... i know how to add proxys curl_setopt($ch, CURLOPT_PROXYTYPE, CURLPROXY_HTTP); curl_setopt($ch, CURLOPT_PROXY,"[i][b]PROXYHERE[/b][/i]"); Here are my questions.. What kind of proxy dose it use? ( Sock5? http? Sock4? ect..) Okay second here the php part. I need it to read a txt file on the server called proxylist.txt and randomly pick a proxy from it but after it picks it. it has to remove it from the TXT file. Help!?@!

instead if cURL using the Servers IP address.. can i make it so it will use the Visitors Ip address? Please show an example of it. Thanks!

Okay, What we have here, is that the script selects a email and password from the database and when it dose it puts used to 0 and then if i vist the page again its suppose to select another one and then set it to 0. But i get this error below if one of the is set to 0 and the script wont use the ones that are not set to 0 Warning: mysql_result() [function.mysql-result]: Unable to jump to row 0 on MySQL result index 2 in index.php on line 16 mysql_select_db($database); $query=mysql_query("select count(*) from where used=0"); $count=mysql_result($query,0); $i=rand(1,$count); //echo $i; $query_email=mysql_query("select email from logins where userID='$i' and used=0"); $email=mysql_result($query_email,0);

Haha okay, But your way wont work.. See the site url is this : https://accounts.site.com/login And below is what needs to be posted to https://accounts.site.com/login in order to login. loginType=L&step=confirmation&[email protected]&inputPassword=test123 So any other idea's? Or maybe yours will work but idk im not that smart LOL.

I need some help! I tried my best but still cant get it, The code Below Will Load the page and put the email in the email box. But it wont Put the password in the password box or submit the form.. So can anyone shed some light and help fix this code? Thanks! <?php // INIT CURL $ch = curl_init(); // SET URL FOR THE POST FORM LOGIN curl_setopt($ch, CURLOPT_URL, 'http://website.com/login.php'); // ENABLE HTTP POST curl_setopt ($ch, CURLOPT_POST, 1); // SET POST PARAMETERS : FORM VALUES FOR EACH FIELD curl_setopt($ch, CURLOPT_PROXYTYPE, CURLPROXY_HTTP); curl_setopt($ch, CURLOPT_PROXY,"http://64.202.165.130:3128"); curl_setopt ($ch, CURLOPT_TIMEOUT, 120); curl_setopt ($ch, CURLOPT_POSTFIELDS, '[email protected]&inputPassword=test123'); // IMITATE CLASSIC BROWSER'S BEHAVIOUR : HANDLE COOKIES curl_setopt ($ch, CURLOPT_COOKIEJAR, 'cookie.txt'); # Setting CURLOPT_RETURNTRANSFER variable to 1 will force cURL # not to print out the results of its query. # Instead, it will return the results as a string return value # from curl_exec() instead of the usual true/false. curl_setopt ($ch, CURLOPT_RETURNTRANSFER, 1); $response = curl_exec ($ch); if(is_int($response)) { die("Errors: " . curl_errno($ch) . " : " . curl_error($ch)); } // CLOSE CURL curl_close ($ch); print "Remote Site : $store<br /><hr />$response"; ?>

Anyone?

Ya i understand but you mixing the 2 problems up.. First problem is have a drop down and i need help getting the information from the post. You know if i have a text box names catname and i hit submit to go to process php i will have $name= $_POST['catname']; so i can echo $name and it will echo what ever was posted.. How do i do that with the drop down box?

Okay I got a drop down box and it grabs tables from mysql and list. now i need to post from it. heres the code <?php $sql = "SHOW TABLES FROM $db_name"; $result = mysql_query($sql); if (!$result) { echo "DB Error, could not list tables\n"; echo 'MySQL Error: ' . mysql_error(); exit; } echo "<select name=Field>"; while ($row= mysql_fetch_row($result)) { echo "<option value={$row[0]}>{$row[0]}</option>"; } echo "</select>"; ?> That works fine but $cat = $_POST['Field']; Dose not work.. Understand? Second problem is my php script to create Tables in mysql works only with single words so if i do Cow it will work but if i do Big Cow i get this error. You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'Cow ( `id` int(255) NOT NULL default '0', `Product` varchar(32) NOT NULL d' at line 1 <? include("conn.php"); if($_POST['submit']) //If submit is hit { $name= $_POST['catname']; // Create a MySQL table in the selected database mysql_query("CREATE TABLE IF NOT EXISTS $name ( `id` int(255) NOT NULL default '0', `Product` varchar(32) NOT NULL default '', `Cat` varchar(32) NOT NULL default '', `Info` text NOT NULL, `Paypal code` text NOT NULL, `Img1` varchar(50) NOT NULL default '', `Img2` varchar(50) default NULL, `Img3` varchar(50) default NULL, `Img4` varchar(50) default NULL, `Img5` varchar(50) default NULL, `Img6` varchar(50) default NULL )") or die(mysql_error()); } header("Location: newcat.php"); exit; ?>

Help list the Mysql Tables in a drop down box? Select Category <? $sql = "SHOW TABLES FROM $db_name"; $result = mysql_query($sql); if (!$result) { echo "DB Error, could not list tables\n"; echo 'MySQL Error: ' . mysql_error(); exit; } while ($row= mysql_fetch_row($result)) { echo "<select name=Field>"; echo "<option value={$row[0]}>"; echo "<option value={$row[0]}>"; echo mysql_table_name($result, $i); } echo "</select></form>"; ?> I know the code is really messed up.. Can someone help.

Okay im trying to get a list of mysql tables from a database to get listed into a html table. Im trying to get a 2 column table like below. Table Names - Link Table 1 Click Me link (la.com/ss.php?table=table1) Table 2 Click Me link (la.com/ss.php?table=table2) Table 3 Click Me link (la.com/ss.php?table=table3) Someone help me do that please! This is what i got but it doesn't work. $sql = "SHOW TABLES FROM $db_name"; $result = mysql_query($sql); if (!$result) { echo "DB Error, could not list tables\n"; echo 'MySQL Error: ' . mysql_error(); exit; } $table = 0; while ($table < mysql_num_rows($result)) { echo "<tr>\n"; echo "<td>\n"; echo "{$row[0]}\n"; echo "</td>\n"; echo "</tr>\n"; } mysql_free_result($result); ?>

Help! Can someone show me how to use Curl and grab the first image off google? So if i have http://www.mysite.com/search.php?q=Car it will grab the first IMAGE off google image search and display it!? I really need this.

Okay this is a date of birth... Drop down menu... I need to pass it in this format dd-m-yr Okay i have 3 drop down menus dd and Month and Yr i need to combin all these into 1 field and make the variable birthdate if you understand. Hpw!?

Well.. If you want to take a look at my code then be my guess. I still don't understand how i can just check those 5 columns. <?php $query = "SELECT count(*) FROM pbs_users WHERE uEmail='".$userInfo[$TbUsers->col_uEmail]."'"); $result = mysql_query($query) or die(mysql_error()); if(empty){ header("Location: info.php"); } }else{ header("Location: index.php"); }