PHP - Php Send Get Variable In Chinese

Hi,

I'm trying to setup a quick PHP script that will grab the email from the url (see below) and after inserting into MySQL db - which is working fine - the script will complete two additional tasks:

1. send that same captured email out to a external db as in shown via http://domain1.com/insert.php?email=$lead (example), but then send to a DIFFERENT source - the originator of the lead - a portback acknowledgement using Header (sending the status and email to http://domain2.com/check.php?e=$lead&s=$status for their records).

See the code below:
-------------------------

Code: [Select]

$lead = $_REQUEST['e_mail'];                // will grab email from posted url string and assign to local variable
$result = mysql_query($command);      // this is just to execute the MySQL insert which works just fine but included here to explain validation below

// Create API Call string to insert lead into iContact folder
$requestURL = "http://domain1.com/insert.php?email=$lead";

// Execute API Call to CAKE
$xml = simplexml_load_file($requestURL) or die("feed not loading");


if ($result) {
    $status = 1;       // mark lead as sucess

    // send postback on lead status
    header("Location: http://domain2.com/check.php?e=$lead&s=$status");
}

--------

Problem: I'm getting all sorts of errors with the simplexml_load_file() function and can't figure out why it won't work.
Any input appreciated as this the only way I know how to pass the lead onward and then inform/update the other party of receipt of information.

thanks!

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]'
         ");}?>

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 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.

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>";
            }

    }

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!

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

 

I am trying the following code to send SMS.


<?php
  $gatewayURL  =   'http://localhost:9333/ozeki?';
  $request = 'login=admin';
  $request .= '&password=abc123';
  $request .= '&action=sendMessage';
  $request .= '&messageType=SMS:TEXT';
  $request .= '&recepient='.urlencode('+9898989898');
  $request .= '&messageData='.urlencode("Hello World");

  $url =  $gatewayURL . $request;

  //Open the URL to send the message
   file($url);
?>


But I am getting the following error message.

Warning: file(http://localhost:9333/ozeki?login=ad...ta=Hello+World) [function.file]: failed to open stream: Connection refused in /home/vpe/public_html/_demo2/sms/sms2.php on line 13

Please help to rectify my code. Thanks a lot in advance.