PHP - If > 1 Then Print 1 Variable.
I currently have a search page on my site that prints the products but it prints the products more than once if its in more than one category I have tried getting distinct item in my SQL. But this doesnt work so im trying an if statement that if there is more than one specific result then to just print this once. I was wondering if anyone had any ideas of how to do this using an if statement I just dont know how to go about just printing the result just once if its greater than 1. The code is below to make it clearer.
$searchterm = $_POST['searchterm']; trim ($searchterm); /*check if search term was entered*/ if (!$searchterm){ echo 'Please enter a search term.'; echo $searchterm; } /*add slashes to search term*/ if (!get_magic_quotes_gpc()) { $searchterm = addslashes($searchterm); } /*query the database*/ $query = "SELECT * from (products LEFT JOIN categories_products_link ON products.prod_id = categories_products_link.prod_id) LEFT JOIN categories ON categories_products_link.cat_id = categories.cat_id WHERE prod_title LIKE '%" . $searchterm . "%' ORDER BY cat_title, prod_title"; $result = mysql_query($query); /*number of rows found*/ $num_results = mysql_num_rows($result); echo '<p><h1>Search Results: '.$num_results.'</h1></p><br />'; /*loops through results*/ for ($i=0; $i <$num_results; $i++) { $num_found = $i + 1; $row = mysql_fetch_assoc($result); echo "$num_found. "?><a href="store-<?php echo $row['cat_id'];?>-<?php echo $row['prod_id']; ?>/<?php echo seo_makeSafeURI($row['prod_title']); ?>.html"><strong><?php echo $row['prod_title']; ?></strong></a> <br /> Similar TutorialsI currently have a problem where I'm trying to list alot of products with the category heading only being returned once and then the long list of products within that category being listed. For Example Football boots Nike Addidas Puma etc Whereas at the moment I'm getting. Football boots Nike Football boots Addidas Football boots Puma My current code for this is as follows. <?php while($row = $db->fetchrow($stuff)) { ?> <li><a href="store-<?php echo $row['cat_id']; ?>/<?php echo seo_makeSafeURI($row['cat_title']); ?>.html"><strong><?php echo stripslashes(htmlentities($row['cat_title'])); ?></strong></a></li> <li><a href="store-<?php echo $row['cat_id']; ?>-<?php echo $row['prod_id']; ?>/<?php echo seo_makeSafeURI($row['prod_title']); ?>.html"><?php echo stripslashes(htmlentities($row['prod_title'])); ?></a></li> <?php } ?> Any help would be much appreciated. I have a class that I created, halfway down my page I call a function from the class which prints some code and also changes a variable inside the class. It's a search class basically it prints out the search results and saves the total amount of results into the variable. I want to display the amount of results above the results like so:- echo $Search->results; $Search->GetResults('foo'); The problem is that $Search->results will be equal to NULL until the variable is updated in the function GetResults. Hope that makes sense, am I going about this wrong? Hello all. I am very new to PHP, and I am not sure where to look or what I'm looking for in my current assignment. My task is to take in two numbers between 0-100. Once I take in that number, it should state beside it "The __ was accepted." The program should not accept any numbers greater than 100 or any characters. Once I do this, I must take a second number and do a similar thing. Finally, I must have a statement show up at the bottom stating which number is greater. Essentially, I need help in determining what I should use to place parameters, and how I can keep the program from echo ing any statement until input has been taken and tested for parameters. Any help you can provide will be greatly appreciated! I am trying to allow the user to update a variable he chooses by radio buttons, which they will then input text into a box, and submit, to change some attributes. I really need some help here. It works just fine until I add the second layer of variables on top of it, and I can't find the answer to this question anywhere. <?PHP require('connect.php'); ?> <form action ='' method='post'> <select name="id"> <?php $extract = mysql_query("SELECT * FROM cars"); while($row=mysql_fetch_assoc($extract)){ $id = $row['id']; $make= $row['make']; $model= $row['model']; $year= $row['year']; $color= $row['color']; echo "<option value=$id>$color $year $make $model</option> ";}?> </select> Which attribute would you like to change?<br /> <input type="radio" name="getchanged" value="make"/>Make<br /> <input type="radio" name="getchanged" value="model"/>Model<br /> <input type="radio" name="getchanged" value="year" />Year<br /> <input type="radio" name="getchanged" value="color" />Color<br /><br /> <br /><input type='text' value='' name='tochange'> <input type='submit' value='Change' name='submit'> </form> //This is where I need help... <?PHP if(isset($_POST['submit'])&&($_POST['tochange'])){ mysql_query(" UPDATE cars SET '$_POST[getchanged]'='$_POST[tochange]' where id = '$_POST[id]' ");}?> My login script stores the user's login name as $_SESSION[ 'name'] on login. For some unapparent reason, i'm getting errors stating that $user and $priv are undefined variables, though I've attempted to define $user as being equal to $_SESSION['name'], using $user to look up the the user's privilege level (stored as the su column ) in the SQL table, and then where the result of the sql query is $priv which is then evaluated in an if statement. I can't seem to figure out why this might not be working. The code I'm using: <?php session_start(); function verify() { //verify that the user is logged in via the login page. Session_start has already been called. if (!isset($_SESSION['loggedin'])) { header('Location: /index.html'); exit; } //if user is logged in, we then lookup necessary privleges. $_SESSION['name'] was written with the login name upon login. Privleges // are written in db as a single-digit integer of of 0 for users, 1 for administrators, and 2 for special users. $user === $_SESSION['name']; //Connect to Databse $link = mysqli_connect("127.0.0.1", "database user", "password", "database"); if (!$link) { echo "Error: Unable to connect to MySQL." . PHP_EOL; echo "Debugging errno: " . mysqli_connect_errno() . PHP_EOL; echo "Debugging error: " . mysqli_connect_error() . PHP_EOL; exit; } //SQL Statement to lookup privlege information. if ($result = mysqli_query($link, "SELECT su FROM accounts WHERE username = $user", MYSQLI_STORE_RESULT)) { //LOOP TO CYCLE THROUGH SQL RESULTS AND STORE Privlege information as vairable $priv. while ($row = $result->fetch_assoc()) { $priv === $row["su"]; } } // close SQL connection. mysqli_close($link); // Verify privleges and take action. Only a privlege of "1" is allowed to view this page. A privlege of "2" indicates special //accounts used in other scripts that have certain indermediate additional functions, but are not trusted administrators. if ($priv !== 1) { echo $_SESSION['name']; echo "you have privlege level of $priv"; echo "<br>"; echo 'Your account does not have the privleges necessary to view this page'; exit; } } verify(); ?>
I have just re-installed Xampp and suddenly my sites are now displaying lots of: Notice: Use of undefined constant name - assumed 'name' in ... Notice: Use of undefined constant price - assumed 'price' in ... this is an example of the line its refering too: $defineProducts[1001] = array(name=>'This is a product', price=>123); I have a form that creates rows of data input textboxes depending on a user input number of things. I have a naming convention for all these textboxes that basically just keeps incrementing a number suffix for each row. All this is working fine. My problem is I need to get the data inserted into this table of textboxes into an array. Here's my code where I attempt to to this (it does not work): Code: [Select] $temp = $_SESSION['Num_Part']; $count = 1; while ($count <= $temp){ $temp2[$count] = "'Participant_P".$count."'"; //echo $temp2[$count]."<br/>"; $temp3[$count]=$_POST[$temp2[$count]]; //here's the problem $temp4[$count] = "'Result_P".$count."'"; $temp5[$count]=$_POST[$temp4[$count]]; //here's the problem //echo $temp4[$count]."<br/>"; $count++; } The problem is that the $_POST does not work with the variable in the argument position - even though the argument is formatted with single quotes. Can a variable be used in a POST argument and if so what is the correct syntax? If not, is there some other simple solution to harvest the data into an array. I understand I can harvest by explicitly accessing each key in the post assoc array. But this could be dozens of rows of input fields. Thanks in advance for your help here. I couldn't find anything online re this topic. I have a script that adds points together based upon the placing. This is the actual script: Code: [Select] <? $points = 0; if($place === '1st') {$points = $points + 50;} elseif($place === '2nd') {$points = $points + 45;} elseif($place === '3rd') {$points = $points + 40;} elseif($place === '4th') {$points = $points + 35;} elseif($place === '5th') {$points = $points + 30;} elseif($place === '6th') {$points = $points + 25;} elseif($place === '7th') {$points = $points + 20;} elseif($place === '8th') {$points = $points + 10;} elseif($place === '9th') {$points = $points + 10;} elseif($place === '10th') {$points = $points + 10;} elseif($place === 'CH') {$points = $points + 50;} elseif($place === 'RCH') {$points = $points + 40;} elseif($place === 'TT') {$points = $points + 30;} elseif($place === 'T5') {$points = $points + 30;} elseif($place === 'Champion') {$points = $points + 50;} elseif($place === 'Reserve Champion') {$points = $points + 40;} echo "Total HF Points: $points"; ?>What it *should* do (my friend's script works the same way and it works) it starts at points = 0, than if there is a first place, it adds 50, and so forth until it reaches the end. It is included into a file, in this area: Code: [Select] <div class="tabbertab"> <h2>Records</h2> <? $query92 = "SELECT * FROM THISTABLE WHERE VARIABLE='$id' OR VARIABLE = '$name' ORDER BY ABS(VARIABLE), VARIABLE"; $result92 = mysql_query($query92) or die (mysql_error()); echo "<table class='record'> <tr><th>Show</th> <th>Class</th> <th>Place</th></tr> "; while($row92 = mysql_fetch_array($result92)) { $class = $row92['class']; $place = $row92['place']; $entries = $row92['entries']; $race = $row92['show']; $purse = number_format($row92['purse'],2); echo "<tr><td>$race</td> <td>$class</td> <td>$place</td></tr>"; } ?> <tr><td colspan='3'><div align='right'><? include('includes/points.php'); ?></div></td></tr> </table> </div> This is the code that is relevant. When ended here, it echoes the last place that appears in the results (such as a 5th place echoing 30 points). When I move it to be included in the while loop, it shows Total Points: 50 Total Points: 25 Total Points: 10 (depending on the results displayed on that page). What am I doing wrong? Hello everyone, I can get Test 2 to successfully operate the if statement using a variable variable. But when I try the same method using a session variable (Test 1) the if statement is not executed. Please could you tell me why the if statement in Test 1 is not being executed? Code: [Select] <?php # TEST 1 $_SESSION[test_variable] = "abcd"; $session_variable_name = "_SESSION[test_variable]"; if ($$session_variable_name == "abcd") { echo "<br>line 373, abcd<br>"; } # TEST 2 $test_variable = "efgh"; $test_variable_name = "test_variable"; if ($$test_variable_name == "efgh") { echo "<br>line 379, efgh<br>"; } ?> Many thanks, Stu Hello.
i am totally new to php and just started to learn now. i just dont understand why the following code is not printing the username that i enter on the page.
Please note that the code itself is saved with the name "basicForm.php".
Thanks.
<html> Code: [Select] $query ="SELECT oneID FROM table WHERE table.PersonID = 'game.PlayerA'" ; $result = mysql_query($query); $row = mysql_fetch_array($result); $oneID = $row[0]; [code] If I then echo "$oneID" why does it not print anything? $result echos resource7 Hello, How to customize print page in php ? I had a page, but I need to print it like an invoice look page. Thanks in advance hi all, I have an language pack for example: languages/en.php: Code: [Select] $en['mail']['letter closing'] = "regards,\n your friend!"; and in my config: Code: [Select] $language = "en"; $include_language = @include("languages/".$language.".php"); if(!($include_language)) { $try_default_language = @include("languages/nl.php"); if(!($try_default_language)) { echo "kan de taalpakket niet vinden<br>"; echo "Could not find the language pack.<br>"; echo "example on error: ".$test." shows nothing"; exit; } } In my function I want to include the language pack for example i have $language = 'en' so I want to include $en['general']['letter closing'] I will do this: Code: [Select] global $language,${$language}['general']; But that gives an error unexpected '[' blah blah. How can i call the variable variable array in the valid php way? I just moved my code from Appserv to EasyPHP and it gave me this error, it was working fine on Appserv...what's with easyPHP ?? I have a function that get's a quick single item from a query: function gimme($sql) { global $mysqli; global $mytable; global $sid; $query = "SELECT ".$sql." FROM ".$mytable." WHERE sid = ".$sid; $result = $mysqli->query($query); $value = $result->fetch_array(MYSQLI_NUM); $$sql = is_array($value) ? $value[0] : ""; return $$sql; // this is what I've tried so far $result->close(); } It works great as: echo(gimme("name")); Then I realized that I could use that as a variable ('$name' in this case) elsewhere. However, I can't figure out how get that new, variable variable 'outside' of the function. As such, echo($name); isn't working outside the function. Is there a way to return a variable variable? In other words, is there a way to make a function that creates a variable variable that will available outside of the function?
Thanks
Quote i need to store a variable from database like if i have "copies" in one of my column in my database then i have to store a particular value for copies store it to $copies here i want that i can store value of copies into $copies $update_book="update book set copies=copies-1 where bookid='$bookid'"; $result=mysql_query($update_book,$linkID1); if($result) { print "<html><body background=\"header.jpg\"> <p>book successfully subtracted from database</p></body></html>"; } else { print "<html><body background=\"header.jpg\"> <p>problem occured</p></body></html>"; } } Probably something simple but I have searched high and low and can't figure this one out. I have a variable that is of the datetime format. I have another variable that is of the time format. I need to add them together. Example: $var1 = 2012-02-24 06:38:22 $var2 = 02:00:00 $var3 = $var1 + $var2 = 2012-02-24 08:38:22 Thanks for the help! Hi, I am trying to get the form field to echo a php variable. The problem is that the form is 'Printed' via php. print "<td width=\"300\" valign=\"top\"><input type=\"text\" name=\"usr\" value=\"i.e. JBloggs \" onfocus=\"if(!this._haschanged){this.value=''};this._haschanged=true;\" tabindex=\"1\"></td>"; Instead of the value being i.e. JBloggs have it echo the $name variable. Any input would be appreciated. This topic has been moved to Installation in Windows. http://www.phpfreaks.com/forums/index.php?topic=351150.0 Probably a simple solution, just one I'm not sure how to do it. I need the array to print out like this Code: [Select] array( 'name'=>'Store 1', 'address'=>'LA1' ), Right now, I got it to look like this Thanks in advance Code: [Select] Array ( [name] => 'Albertville Farmers Market', [address] => '116 Main Street Albertville, Alabama 35950' ) , Array ( [name] => 'Alexander City Downtown Market', [address] => 'Braod Street Alexander City, Alabama 35010' ) , How can I go about accomplishing this? Here is my code Code: [Select] $stores = array('name'=>"'$MktName',",'address'=>"'$address'"); echo "<pre>"; print_r($stores); echo "</pre>"; |