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PHP - All Script In One Variable! Possible?
I have this script and i want to take this in one variable & print where i want is this possible
Code: [Select] if ($nro > 0) { echo "<table id='gridview' align='center'> <tr> <td bgcolor='#99CC00'>E.T.D.</td> <td bgcolor='#99CC00'>E.T.A.</td> <td bgcolor='#99CC00'>Quantity</td> </tr>"; $totqty=0; while ($row=mysql_fetch_array($result)) { $etd = $row['etd']; $etd = GetDateUserFormat($etd); $eta = $row['eta']; $eta = GetDateUserFormat($eta); $qty = $row['shptqty']; echo "<tr>"; echo "<td>" . $etd . "</td>"; echo "<td>" . $eta . "</td>"; echo "<td>" . $qty . "</td>"; echo "</tr>"; $totqty = $totqty + $qty; } $remqty = $styleqty - $totqty; echo "<br>"; echo "<tr>"; echo "<td bgcolor='#99CC00' colspan='2'> Total Entered Quantity</td>"; echo "<td>" . $totqty . " Kg.</td>"; echo "<td bgcolor='#99CC00'> Remaining Quantity</td>"; echo "<td>" . $remqty . " Kg.</td>"; echo "</tr>"; echo "</table>"; } Similar TutorialsI 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, I have this script, it all works fine apart from when I pass the variable on, it only passes the first word and not the whole variable The form that actions to the samples page passes the $siteName as My Test Site, if I type echo $siteName in the sameples page, it will print My Test Site, however when I use this Code: [Select] createSample&siteName=$siteName to pass the variable to the next page, it just echos 'My' instead of My Test Site. Even when I scroll over the link above it just shows My in the url. Code: [Select] case "samples": // Get submitted data and assign to variables $siteName = $_POST['siteName']; $adminEmail = $_POST['adminEmail']; $sendmailLoc = $_POST['sendmailLoc']; $imgdir = $_POST['imgdir']; $imgdirbase = $_POST['imgdirbase']; // Write variable data to text file $FileName = "data/server.txt"; $FileHandle = fopen($FileName, 'w') or die("can't open file"); $data = "<?php\n\$siteName = '$siteName';\n?>"; fwrite($FileHandle, $data); $data = "<?php\n\$adminEmail = '$adminEmail';\n?>"; fwrite($FileHandle, $data); $data = "<?php\n\$sendmailLoc = '$sendmailLoc';\n?>"; fwrite($FileHandle, $data); $data = "<?php\n\$imgdir = '$imgdir';\n?>"; fwrite($FileHandle, $data); $data = "<?php\n\$imgdirbase = '$imgdirbase';\n?>"; fwrite($FileHandle, $data); fclose($FileHandle); // Displays the create sample pages page ECHO <<<SAMPLES <table border=0 align=center bgcolor=#00CCFF> <tr> <td><span class=style1><b><center>Create Sample Pages</center></b></span></td> </tr> <tr> <td>Would you like Member Site Maker to create sample pages for: <ul> <li>index.html</li> <li>login.html</li> <li>search.html</li> <li>register.html</li> </ul> </td> </tr> <tr> <td><a href=installation.php?cmd=createSample&siteName=$siteName>Yes</a> </td> <td><a href=installation.php?cmd=mysql>No</a> </td> </tr> </table> SAMPLES; break; Can anybody tell me where I am going wrong please? Thanks basically if you read this, then you can see they have a "passing-var.php" and a "catching-var.php" How can i make this happen without the passing-var.php and make the catching-var.php actually GET a variable How can i put an variable for example
$variable = 'something';
and then how do i put this into this js script
<script>$(document).ready(function() {$('#open$variable').dataTable();});</script>
note: Im trying to put this '$variable' inside ' ' right after #open.
Attempting to set a variable = NOW() + INTERVAL 90 DAY; I can run SELECT DATE_ADD(NOW(), INTERVAL 90 DAY using MYSQL, but don't know how to move result to a variable. 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; } ?> hello everybody! This is just basic script, where i try to modify for the needs. I try to play with it. i want to parse some data. The whole script has three parts: 1. Fetching 2. parsing 3. storing i want to put all into one script: Two are allready put together - there all seems to be clear... So this thread is one that asks for the combining two parts of a script - how to invoke a variable between them What has happened untill now: 1st i need to have a connection to database lets say MySQL. I will suggest to use mysqli instead of mysql. Well - okay i safe this db.php $host="localhost"; //database hostname $username="******"; //database username $password="******"; //database password $database="******"; //database name ?> Now i am going to take a new script and save this config.php <?php require_once('db.php'); //call db.php $connect=mysqli_connect($host,$username,$password); //connect to mysql through mysqli if(!$connect){ die("Cannot connect to host, please try later."); //throw error if any problem } else { $select_db=mysqli_select_db($database); //select database if(!$select_db){ die("Site Database is down at the moment, Please check later. We will be back shortly."); // error if cannot connect to database or db does not exist } } ?> Now i have to take care for the script, that takes the files (note this is very basic - it is only a proof of concept. In the real situation i will take cURL since cURL is much much nicer and more elegant and faster etc. <?php require_once('config.php'); // call config.php for db connection $content = file_get_contents("<-here the path to the file goes in-> Position XY! an URL is here "); var_dump($content); $pattern = '/<td>(.*?)<\/td>/si'; preg_match_all($pattern,$content,$matches); foreach ($matches[1] as $match) { $match = strip_tags($match); $match = trim($match); var_dump($match); $sql = mysqli_query("insert into tablename(contents) values ('$match')"); } ?> Note: This is just basic script, where you can modify it for your taste and can play with it. Question: If i have stored the URLs that i want to parse in a local file - how do i "call" them in the script. How do i do the call to file where the URLs (there are more than 2500 URLs that have to be parsed) at the following position: $content = file_get_contents("<-here the path to the file goes in-> Position XY! an URL is here "); The folder with the URLs is stored in the same folder as the scripts reside! Many thanks for all hints and for a starting point! if i have to write more - or if you need more infos - or if i have to be more concrete, just let me know! i love to hear from you! db1 Hi
I am trying to create a google chart from data in query.php and using the google api to load it. Everything works until I want to change the MetalSourceID from a drop select box.
PHP CODE FOR DROP DOWN BOX:
<form>
<select name="users" onchange="showUser(this.value);drawChart();">
<option value=""> Select a Metal: </option>
<?php
$query = "SELECT TOP(31) tblMetalPrice.MetalSourceID, tblMetalSource.MetalSourceName from tblMetalPrice INNER JOIN tblMetalSource ON tblMetalPrice.MetalSourceID=tblMetalSource.MetalSourceID ORDER BY tblMetalPrice.DateCreated DESC ";
$result = sqlsrv_query( $conn, $query);
while( $row = sqlsrv_fetch_object ($result)) {
echo "<option value='".$row->MetalSourceID ."'>". $row->MetalSourceName ."</option>";
}
sqlsrv_close( $conn);
?>
</select>
</form>
This works fine and generates all the correct values. One part of this changes contents of a table which works fine. But I also echo the MetalSourceID into the javascript for the google api, JS script below:
<script type="text/javascript">
google.load('visualization', '1', {'packages':['corechart']});
google.setOnLoadCallback(drawChart);
function drawChart() {
var jsonData = $.ajax({
url: "query.php",
dataType:"json",
async: false,
data: { 'MetalSourceID' : <?php $q = intval($_GET['q']); echo $q; ?> }
}).responseText;
var data = new google.visualization.DataTable(jsonData);
var chart = new google.visualization.LineChart(document.getElementById('chart_div'));
chart.draw(data);
}
</script>
This then runs the query.php script and returns the google line chart, a copy of the query.php script is below:
I have a php string variable that is created by php code within an html form ($answer). I need to pass this string variable along with all the html form input data to another php script specified with the form "action" (post method). All the html form input data is coming thru fine but not the variable ($answer). How do I do this? Here is the php code for importing html form data at the script called in the form action: $languages = $_POST['languages']; $answergiven = $_POST['answergiven']; $problemanswer = $_POST['$answer']; 'languages' and 'answergiven' are form inputs and come thru fine. '$answer' does not get passed to the second script. How do I do this? Here is the php code within the first html form <?php // OPEN DATABASE $username="servics3_sample"; $password="sample"; $database="servics3_sample"; mysql_connect(localhost,$username,$password); @mysql_select_db($database) or die( "Unable to select database"); // GENERATE RANDOM PROBLEM NUMBER $probnum = (rand ( 1 , 9 )); echo $probnum; // RETRIEVE ANTI-SPAM PROBLEM $query="SELECT * FROM liasantispam WHERE `problem number` LIKE '%$probnum%' "; $result=mysql_query($query) or die("Error: ". mysql_error(). " with query ". $query); $firstnum=mysql_result($result,0,"first number"); $operator=mysql_result($result,0,"operator"); $secondnum=mysql_result($result,0,"second number"); $answer=mysql_result($result,0,"answer"); echo $firstnum," ",$operator," ",$secondnum," = "; mysql_close(); ?> Hi guys, Basically I am developing a page where the order status of customers are listed in a tabular form, what I need now is to mail these data to my email address at a defined date (e.g. 1st day of next month) automatically. I do not know how to code this since I am just self studying and is still a noob. Thanks. Hi all, Thanks for reading. I'm hella frustrated at this script I wrote: for some reason, it will not work correctly. Basically, it works. The first 4 names in the table on the database show up when searched. But, anything past these four names in the database will not show up as a result when searched! I'm pulling my hair out here! It's really simple - take a gander: Code: [Select] if (isset($_POST['submit'])) { $search = $_POST['search']; $searchQuery = mysql_query("SELECT * FROM Accounts WHERE FullName='$search'"); if (mysql_num_rows($searchQuery) == 0) { $result = "Your search returned no results. Please try again."; } else { $results = 1; while ($getSearchResults = mysql_fetch_array($searchQuery)) { $fullName = $getSearchResults['FullName']; $result = "Name: ".$fullName.""; } } } ?> ...and the HTML form... Code: [Select] <form action="search.php" method="post"> <p>Search: <input type="text" name="search" size="35" maxlength="100" /></p> <p><input type="submit" value="Search" name="submit" /></p> <?php echo $result; ?> </form> Does anyone have any ideas? 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();
?>
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 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? 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! |
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