PHP - Can A Variable Be Changed To Another Variable
A silly question, since simply fixing the error would likely take less time, but my curiosity won't stop nagging me. I have several php pages that ALL use the INCLUDE file connection.php which has the following scripting $conn = mysqli_connect(...etc, etc,etc,); Everything works fine EXCEPT for a few pages which were written using $DBCONN instead of $conn Can I somehow alter the value of $DBCONN so it is equal to $conn And interpreted correctly without having to change the mistaken variable in the pages that use $DBCONN ? I tried a quick $DBCONN = $conn; but there was no effect, so I thought I'd ask.
Similar TutorialsHello 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]' ");}?> 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); 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 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. 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 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? 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 $language,${$language}['general']; But that gives an error unexpected '[' blah blah. How can i call the variable variable array in the valid php way? I have a function that get's a quick single item from a query: function gimme($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>"; } } 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 ?? 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! Code: [Select] $word = 'numbers'; $numbers= array('1', '2', '3', '4'); echo $$word[0]; I expected the output to be '1'. It ended up being nothing... Why does this not work? Is it not possible to have a variable variable array? And if not, is there a workaround? Cheers, Joe I'm trying to loop through a database and assign the results to a variable that is created with the loop with the variables name based on which instance is currently running in the loop. I'll go ahead and include the code I thought might work, but I obviously have no idea what I'm doing in this case $i = 0; while ($i < $num_results) { $ip = ($i++); $result = ("$res_" . "$ip"); $loop_date = ("$res_" . "$ip" . "_date"); $get_result = "SELECT * FROM table WHERE id = '$result'"; $que_result = mysql_query($get_display); while ($sub_res = mysql_fetch_array($que_result, MYSQL_BOTH)) { $loop_date = $sub_res['date']; } $i++; } echo $res_1_date; Can anyone please clue me in on how I can get this to work please. It works with $result but it isn't working with the last $res_1_date... As I'm writing this I see where the problem is but just not quite sure how to make it work... Hi, I have following code : Code: [Select] <?php class CategoryWork { public $form = ""; public $error = ""; public $add_form = "<br /><br /><p><strong>Add New Category</strong></p><br /><form id=\"form1\" name=\"form1\" method=\"post\" action=\"category.php?action=add\"> <table width=\"550\" height=\"170\" border=\"0\"> <tr> <td width=\"153\">Name :</td> <td colspan=\"2\"><label for=\"cat_name\"></label> <input name=\"cat_name\" type=\"text\" id=\"cat_name\" size=\"50\" maxlength=\"50\" /></td> </tr> <tr> <td>Slug :</td> <td colspan=\"2\"><label for=\"cat_slug\"></label> <input name=\"cat_slug\" type=\"text\" id=\"cat_slug\" size=\"50\" maxlength=\"50\" /></td> </tr> <tr> <td>Description</td> <td colspan=\"2\"><label for=\"cat_desc\"></label> <textarea name=\"cat_desc\" id=\"cat_desc\" cols=\"48\" rows=\"10\"></textarea></td> </tr> <tr> <td> </td> <td width=\"97\"><input type=\"submit\" name=\"button\" id=\"button\" value=\"Submit\" /></td> <td width=\"286\"><input type=\"reset\" name=\"button2\" id=\"button2\" value=\"Reset\" /></td> </tr> <tr> <td> </td> <td colspan=\"2\">error</td> </tr> </table> </form>"; } ?> At the line of <td colspan=\"2\">error</td> I want to insert $error, hence I made the line as <td colspan=\"2\">".$error."</td> and also tried <td colspan=\"2\">".$this->error."</td> After executing with this, php says syntax error at the line of public $add_form I am going to declare an error at the $error When I do <td colspan=\"2\">$error</td> it says : Parse error: syntax error, unexpected '"' at line of public $add_form When I do <td colspan=\"2\">".$error."</td> it says : Parse error: syntax error, unexpected '.', expecting ',' or ';' at the above line i.e. of <td> I am damn confused, I googled but nothing found any helpful material. Anybody please help me.. Thanks Hi, so I have an easy problem that for some reason I couldn't find an answer to anywhere. I have a bunch of variables like this: $pic1fileName $pic2fileName $pic3fileName $pic4fileName $pic5fileName...you get the idea So I have another variable I'm pulling from a database that specifies which number to show, so I need a variable something like this: $pic($pic_number)fileName I just don't know what the proper syntax is. Anyone? Thank you freaking much for any help; this is a lame problem. I declared a variable in a page. When i want to access that variable in other page i am bound to state as global such as global $var; now i can not access this variable in other pages even if declared global. Please some one tell me how to declare a variable once and can use in any page i want. I have a session variable that will be set to a number called... Code: [Select] $_SESSION['number'] I have page where I want to set a variable in the format of $serv# (with "#" being the value of the session variable). Is there a way to write one simple line of code that in effect says... Code: [Select] $serv . $_SESSION['number'] = "selected='selected'"; I'm just looking to avoid having to write this code... Code: [Select] if ($_SESSION['number'] == 1) { $serv1 = "selected='selected'"; } elseif ($_SESSION['number'] == 2) { $serv2 = "selected='selected'"; } etc etc It's basically putting a variable inside of a variable and I'm not sure if this is allowed/proper? Any insight would be appreciated. Thanks, Gary Hi i have this scroolbar javascript where you implement the text inside the java variable but because i want the data to come from the database i wanted to use a php example <? print($marker); ?> so i was wondring if it is possible to add a php variable into a javascript like this my javascript at moment the javascript variable is like this Code: [Select] var myMainMessage=" Try this out! The message just keeps repeating. Use it for announcements, news and other items! ";i want to change to this Code: [Select] var myMainMessage="<? print($marker); ?>";is it possible if so can someone show me an example thanks the whole javascript Code: [Select] <SCRIPT LANGUAGE="JavaScript"> <!-- This script and many more are available free online at --> <!-- The JavaScript Source!! http://javascript.internet.com --> <!-- (c) http://www.wyka-warzecha.com --> <!-- Begin // THESE VARIABLES where i want to add the php // var myMainMessage=" Try this out! The message just keeps repeating. Use it for announcements, news and other items! "; var speed=150; var scrollingRegion=50; // END CHANGEABLE VARIABLES // var startPosition=0; function mainTextScroller() { var mainMessage=myMainMessage; var tempLoc=(scrollingRegion*3/mainMessage.length)+1; if (tempLoc<1) {tempLoc=1} var counter; for(counter=0;counter<=tempLoc;counter++) mainMessage+=mainMessage; document.mainForm.mainTextScroller.value=mainMessage.substring(startPosition,startPosition+scrollingRegion); startPosition++; if(startPosition>scrollingRegion) startPosition=0; setTimeout("mainTextScroller()",speed); } // End --> </script> So I'm fairly new to php and just need some quick help. Here is my code: Code: [Select] function plaintext_category(){ $cat_plain = strip_tags( get_the_term_list($post->ID, 'portfolio_category', '', ', ', '' ) ); $cat_plain = strtolower($cat_plain); $cat_plain = str_replace(' ','-',$cat_plain); echo $cat_plain; } $pattern = '/title=\"(.*?)\"/'; $replace = 'title="echo $cat_plain"'; /* This is where i need help, how do i echo this? */ $categories = preg_replace($pattern,$replace,$categories); echo $categories; So I'm trying to pass the echo value of $cat_plain but when I put echo in front of it, i get an error. If I don't put echo, I just get a blank result. Please, need some help on this. Thanks guys! |