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PHP - External Variable For Url
Hi all,
So im trying to improve on my PHP as my knowledge isn't that great.
lets say for argument sake my ip is:
192.168.0.1
im trying to set a external variable to set this as the url.
so on each page i use, for each link instead of typing out the ip address i can simply type:
<a href="{url}/index.php">Home</a>
and then if it is possible, to do something like this:
<link rel="stylesheet" href="{url}/main.css" type="text/css">
all in one file so in my php pages i can simple include 1 php file and it will have all of the relevant stylesheets ect linked to it.
Edited by srwright, 30 May 2014 - 01:18 PM. Similar Tutorialswhen tyring to pass a variable to a extenal javascript file it just shows up as a sting and the variable isnt beening executed
var timezone_offset = <?= $timezone_offset ?>;im guessing its probably down to the fact the variable is being passed from my framework controller... so im wondering how do you send a variable to a external javascript file when using a framework?
$this->view->timezone_offset = "test"; // in my controller
// passing varibales to template in my view
if (preg_match("/\.html$/i", $file_name))
{
extract($this->_variables);
require_once PRIVATE_DIRECTORY . 'application' . DS . 'views' . DS . $file_name;
}
thanks guys
Hey, I am currently trying to get a variable created inside a require_once script to be echoed inside the main page that called the require. The script below is a basic idea of what i want to do. I just want to be able to create a basic variable none of this session stuff as its makes life harder at the moment. Thank guys, hope the snippet below gives you a better idea. Main Code: Code: [Select] <body> <?php require_once("makesVariable.php"); <div> // Variable I want to be echo "NOT WORKING" echo $var; </div> ?> </body> External PHP Code: Code: [Select] <?php //Function gets called by previous code to create the needed variable function createTheVariable(){ $var = "I am the variable to be called"; return $var; } ?> I'm trying to send a submission to the following Craigslist form that uses javascript to trigger an auto submit. The user selects "For Sale" on my form and it will pass the data in a hidden div to the proper selection on the craigslist page. Heres the page it would be passing to. I know I will have to pass it in the URL. Code: [Select] <!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd"> <html> <head> <base href="https://post.craigslist.org"> <title>houston craigslist | choose type</title> <meta http-equiv="Content-Type" content="text/html; charset=ISO-8859-1"> <link type="text/css" rel="stylesheet" media="all" href="/styles/craigslist.css?v=9"> </head> <body id="pp"> <table width="100%" id="header" summary="header"> <tr valign="top"> <td><a href="http://houston.craigslist.org/"><b>houston craigslist</b></a> > choose type<br></td> <td width="10%" class="highlight" style="text-align: right; white-space: nowrap;"> <font size="2" face="sans-serif" color="#7a7a7a">[ logged in as <a href="https://accounts.craigslist.org/login"><b>kennymahaffey@gmail.com</b></a> ] [ <a href="https://accounts.craigslist.org/logout">logout</a> ]</font> <br></td> </tr> </table> <hr> <blockquote> <div class="highlight"> <i>Please post to a single geographic area and category only -- cross-posting to multiple cities or categories is not allowed</i> </div> <h4>What type of posting is this:</h4> <form action="https://post.craigslist.org/k/kInfxI4M4RGIA7PWTF5SWg/HjdlJ" method="POST"> <blockquote> <label> <input type="radio" name="id" value="jo" onclick="form.submit(); return false;">job offered </label> <br> <label> <input type="radio" name="id" value="jw" onclick="form.submit(); return false;">resume / job wanted </label> <br> <br> <label> <input type="radio" name="id" value="ho" onclick="form.submit(); return false;">housing offered </label> <br> <label> <input type="radio" name="id" value="hw" onclick="form.submit(); return false;">housing wanted </label> <br> <br> <label> <input type="radio" name="id" value="fs" onclick="form.submit(); return false;">for sale </label> <i>(please do not post prohibited <sup><a target="_blank" href="http://www.craigslist.org/about/prohibited.items">[?]</a></sup> or recalled <sup><a target="_blank" href="http://www.craigslist.org/about/recalled_items">[?]</a></sup> items)</i> <br> <label> <input type="radio" name="id" value="iw" onclick="form.submit(); return false;">item wanted </label> <br> <br> <label> <input type="radio" name="id" value="go" onclick="form.submit(); return false;">gig offered </label> <i>(I'm hiring for a short-term, small or odd job)</i> <br> <label> <input type="radio" name="id" value="so" onclick="form.submit(); return false;">service offered </label> <br> <br> <label> <input type="radio" name="id" value="p" onclick="form.submit(); return false;">personal / romance </label> <br> <br> <label> <input type="radio" name="id" value="c" onclick="form.submit(); return false;">community </label> <br> <label> <input type="radio" name="id" value="e" onclick="form.submit(); return false;">event </label> <br> <br> </blockquote> <input type="hidden" name="U2FsdGVkX:18yNDg2NDI0O:A8dZroeLa8K5Y677RwK4hzBe2OPdD3XxLUDb5lS9LllELhxNpMqXZIwAWDxJ.7Wo4A" value="U2FsdGVkX18yNDg2NDI0OFjS1qgWn_NKgmgnh5qCZ9aK2m7eOmS-uAAM_Pwu8VHN"> <button type="submit" name="go" value="Continue">Continue</button> </form> </blockquote> 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]' ");}?> 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 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();
?>
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? 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 would like a part of my script to link to an external sites script It doesn't seem to be doing it though Othersite.com - index.php?name=hello&status=1 I would like my post.php to run that above. Can it be done? 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
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! 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 was just wondering if it was possible to have an external php file that can be included in the head of a web page, like a .js file. If this isn't possible maybe have a .js file containing php code that can be executed regarding the JavaScript around it... I'm building a program in php that will be able to view YouTube videos from the URL. So if I type: http://www.youtube.com/watch?v=(Video ID) Then my program works and it will return the Video ID! But then some idiot clicks on a related video and this URL is generated: http://www.youtube.com/watch?v=(Video ID)&feature=related And my program strips the first bit by replacing "http://www.youtube.com/watch?v=" with an empty string "" but then I'm still left with the "&feature=related" I thought about just replacing that with an empty string as well but sometimes there can be a URL like this: &feature=g-vrec&context=G28d9eecRVAAAAAAAABA Which has a different unique code each time. So I thought it'd be much simpler if I could use $_GET[] with an external URL, so the user types in: http://www.youtube.com/watch?v=(Video ID)&feature=related It just gets the "v" value rather than my buggy replace thing. Thanks. Hi, I have a peice of code which publishes an image with a link from my database. However I cant get it to use external links. My code is: echo "<a href=\"" .$link . "\"> <img src=\"" .$image ."\" /> </a><BR />"; I have tried all the options I can think of but I cant get it work. Can anyone advise please? (Posted this in the MySQL section, but i think the php section might be the proper place)
I have a webshop db where i want to export some things linke order information to another external db on another server etc.
Example of things to move:
Table1 - orderid - orderdate Table2 - adress - shippingmethodI ONLY want it to export and import things that is not currently there. How can this be done in a php script? |
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