PHP - Help Setting Variable

Hi

I'm having a problem getting a query to work.

I have a simple form with user input for start and end date with format: 2009-03-19 (todays date):

$Startdate = $_POST['date'];

This works well when something is entered into the form, and afterwards using my query:

SELECT COUNT(*) as total FROM mydb WHERE Date BETWEEN '$Startdate' AND '$EndDate' ........

Problem is if user submits the form without entering anything in the date input fields, which makes sense.

I want to check if inputs has been made, and if not set af default date, but can't make it work:

if (isset($_POST['date']) && $_POST['date'] !='') {
$Startdate = $_POST['date'];}
else { $Startdate = '1980-01-01';}

How can I set $Startdate to something that can be used in the query as below doesn't work?

When I run

'select 1700-price as blah from goldclose as t2 order by dayid desc limit 1'

by itself in mysql I get a numerical result: one row, one column.

In my php script, the 1700 is actually a variable. so here it is

$changequery = sprintf("select $goldprice-price as change from goldclose order by dayid desc limit 1");
$change = mysql_query(changequery);
while ($row = mysql_fetch_array($change)) {
    printf("$row[0]"); 
}

mysql_free_result($changeresult);


I get the following error,

Warning:  mysql_fetch_array(): supplied argument is not a valid MySQL result resource in <b>/root/fuzzy/htmlmain4.php on line 99

Warning:  mysql_free_result(): supplied argument is not a valid MySQL result resource in <b>/root/fuzzy/htmlmain4.php on line 103

Not sure why? All i want is to get the result of that select statement into a variable such as $change

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();
?>

 

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.

can anyone tell me why $username is not setting properly?  the password is, but $username is no.  it is setting as blank:
Code: [Select]
if ($_POST['go']) {
//Connect to the database through our include
if ($_POST['logid'] ==  "&#59;5757120050531338739&#61;49121200000&#63;"){
$username="oldcheney";
$username = stripslashes($username);
$username = strip_tags($username);
$username= strtolower($username);
$username = mysql_real_escape_string($username);

$password="password";}
else if($_POST['logid']=="&#59;5757120050531338721&#61;49121200000&#63;"){
$username="yankeehill";
$password="password";}
else if($_POST['logid']=="%6277202936423578^00000000^X&#63;"){
$username="holdrege";
$password="password";}

$password = md5($password);

I made a website in PHP mvc framework. Now I want to remove "public" keyword from URL. I tried by removing  require "../app/init.php";  these dots and bringing htaccess & index.php file in root directory. But it doesn't work. I searched on google but didn't find anything. Here is my index page code:

 

Quote

<?php 
session_start();

require "../app/init.php";

$app = new App();

 

and my htaccess code:

 

Quote

RewriteEngine On

RewriteCond %{REQUEST_FILENAME} !-f
RewriteCond %{REQUEST_FILENAME} !-d
RewriteRule ^(.*)$ index.php?url=$1 [L,QSA]

Please give me a solution to it. Thank you

I am trying to separate this information by the "-" symbol so that I can insert it.

This output would be ideal.  Any help today?

105-2-2-3-Cheerleading

$value[0] should = 105
$value[1] should= 2
$value[2] should= 2
$value[3] should= 3
$value[4] should= Cheerleading


I have tried many approaches and come up with notta


105-Athletics
105-1-Eagle Sports
105-2-Programs
105-2-1-Boys
105-2-1-1-Football
105-2-1-2-Cross Country
105-2-1-3-Golf
105-2-1-4-Basketball
105-2-1-5-Baseball
105-2-1-6-Track
105-2-1-7-Tennis

105-2-2-Girls
105-2-2-1-Volleyball
105-2-2-2-Golf
105-2-2-3-Cheerleading
105-2-2-4-Basketball
105-2-2-5-Softball
105-2-2-6-Track
105-2-2-7-Tennis

105-3-Coaches
105-4-FHSAA
105-5-Game Directions

Thank you for looking.

hello,

ok if i set a Global on index.php  (wwwroot/admin/index.php) it works

Code: [Select]
$page1='hello from page1 on index';
global $page1;
echo $page1;

i can echo $page1 on other pages like (wwwroot/admin/pages/home.php) and it will echo out "hello from page1 on index".

GREAT

but if i write a Global on (wwwroot/admin/pages/home.php) it does not work. i also can not echo it out on any other page

Code: [Select]
$page2='hello from page2 on home';
global $page2;
echo $page2;



why is this ???

hi
i wants to disable open_base_dir for my folder /home/admin/mysite.com/myfolder/
so in myfolder script i could use CURLOPT_FOLLOWLOCATION

I can disable open_base_dir in my kloxo panel but i would like to use it this way,
I modified my php.ini file and set this path
open_basedir = /home/admin/mysite.com/folder/

but still i get error while using CURLOPT_FOLLOWLOCATION

Warning: curl_setopt() [function.curl-setopt]: CURLOPT_FOLLOWLOCATION cannot be activated when in safe_mode or an open_basedir is set in

Please any help on how to disable that in /myfolder.?
using Centos 5
Thanks

I have created my lottery script.
At the moment i simply have the time() function being used where the draw will happen.
Im wondering if your able to get rid of that and just choose a particular time where i want the draw to happen.
I want the draw to be done at 4pm everyday. (in GMT)

Is this possible? and what would the right code be for it.