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PHP - Genrandomstring But Nothing Displays On Echo
Similar TutorialsOK, have no idea what's going on... I've done this a million times... why wont this output!?? I must have a major brain meltdown and dont know it yet!!! Code: [Select] <?php // this echoes just fine: echo $_POST['testfield']; // but this wont echo: echo if (isset($_POST['testfield'])) { $_POST['testfield'] = $test; } echo $test; /// or even this DOESNT echo either!: $_POST['testfield'] = $test; echo $test; ?> Hi All, I'm trying to echo the response from an SLA query, the query works and returns the data when I test it on an SQL application.. but when I run it on my webpage it won't echo the result. Please help? <?php $mysqli = mysqli_connect("removed", "removed", "removed", "removed"); $sql = "SELECT posts.message FROM posts INNER JOIN threads ON posts.pid=threads.firstpost WHERE threads.firstpost='1'"; $result = mysqli_query($mysqli, $sql); echo {$result['message']}; ?> Hi. I am pretty new to this and am having difficulty looping through rows. I have the code below which first selects a name based upon the id sent from a GET on a separate page. It then has a second select query to get the data from all rows which have this name. I then want it to display the name, and city once but all of the comments. The code below only shows one of each though. Thanks in advance Code: [Select] <?php $id = $_GET['id']; // Retrieve data from database $sql="SELECT landlord_name AS name FROM $tbl_name WHERE id=$id"; $query=mysql_query($sql); //loop through the rows while ($rows = mysql_fetch_array($query)){ $name = $rows['name']; } $sql2="SELECT city, landlord_name, landlord_comments FROM $tbl_name WHERE landlord_name= '$name'"; $query2=mysql_query($sql2) or die(mysql_error()); while ($data = mysql_fetch_array($query2)){ ?> <table width="600" border="1" cellspacing="0" cellpadding="0" align="center"> <tr> <div id="name"><? echo $name ?></div> <div id="city"><? echo $data['city']; ?></div> <div id="comments"><? echo $data['landlord_comments']; ?></div> </tr> </table> <? } //End While ?> I have set up two directories in www which reflect two domains I have using uniform server z which is a wamp package.
I have set up Vhosts also.
The sever is accessible from the internet The directories hold a simple hello world page ( one index HTML and one index.php) I set these up on uniform server z to test access as I have previously had problems getting virtual host to work on coral 6.8 I can test the IP address fine and ping the sever. I can access the two addresses from the net (not on my local network) How ever when I access either site externally I get the Index of/ My DirectoryIndex in the main apache config allows index.php & index.html. I also have that in the .htaccess and in the Vhosts ht access. The Index of however only shows a favicon and not the index.html/php Any help appreciated. Hi There, I have the following code: two arrays: $user_body_group & $DB_Seconday_muscles I want to look up the "$DB_Seconday_muscles" Array and search for the elements in the "$user_body_group" array The result would be displaying a checkbox list with only the items in the "$user_body_group" array checked: foreach($DB_Seconday_muscles as $value) { echo "<input name=\"colors[]\" type=\"checkbox\" value=\"$value\""; if (in_array($value,$user_body_group)) { echo "CHECKED"; } echo "> $value "; } BUT... When i run thi script i only get the first element in the searched array ticked. Can anyone help! Many Thanks! Leatfield I must be mixing apples and oranges here or something trying to get two columns/fields of MySQL data to display.
The basic HTML page display OK and there are no MySQL Connection errors (finally resolved those).
<HTML snipped>
<?php
require '...<URL snipped>...';
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$sql="SELECT `name`, `id` FROM `roster`";
$result = mysqli_query($con,$sql);
$num=mysqli_num_rows($result);
mysqli_close($con);
?>
<table border="0" cellspacing="2" cellpadding="2">
<tr>
<td>
<font face="Arial, Helvetica, sans-serif">NAME</font>
</td>
<td>
<font face="Arial, Helvetica, sans-serif">ID</font>
</td>
</tr>
<?php
function mysqli_result($res, $row, $field=0) {
$res->data_seek($row);
$datarow = $res->fetch_array();
return $datarow[$field];
}
$i=0;
while ($i < $num) {
$f1=mysqli_result($result,$i,$datarow[$field]);
$f2=mysqli_result($result,$i,$datarow[$field]);
?>
<tr>
<td>
<font face="Arial, Helvetica, sans-serif"><?php echo $f1; ?></font>
</td>
<td>
<font face="Arial, Helvetica, sans-serif"><?php echo $f2; ?></font>
</td>
</tr>
<?php
$i++;
}
?>
</table>
<?php
?>
<HTML snipped>
Any assistance is appreciated.
Thanks very much.
- FreakingOUT
Howdy! A client's web page is .shtml and includes various PHP files via standard virtual include: Code: [Select] <!--#include virtual="session.php" --> These PHP includes worked fine until recently, when PHP files that produce no visible output began displaying a 0 (zero). For instance, the session.php file above is here in it's entirety: Code: [Select] <?php session_start(); ?> The include is on the first line of the .shtml file, and includes that bit of code, but displays a 0 in the browser. If I echo anything at all the 0 goes away, but of course we don't want to echo anything before the <html> tag in the .shtml file. The short-term and unacceptable solution is to echo an empty HTML comment, as follows: Code: [Select] <?php session_start(); echo "<!-- -->"; ?> As noted, this causes the 0 to disappear but now we have <!-- --> before the HTML. Repeat: This was not happening when the .shtml and PHP was installed, and the files have not changed. Therefore, it seems that this problem is caused by a change in PHP and/or Apache configuration. Has anyone seen this and/or can anyone explain what's going on? Thanks! First, I would like to say when i tried to recover my account from this website, I took me 10 attempts to get the captcha right and then finally it said it sent me an email to my gmail account. I checked spam folder and everything there was no such email from this site. Then I decided to create a new account, well, it took me another 10 attempts to get the captcha right and finally when it was submitted, the page was loading for around 3 minutes before it signed me in.
My question is about the php date() function. It accepts a format to display a time. In the following example I use F for full representation of month, d for 2-digit day of month with leading zeros, Y for full year, g for 12-hour format without leading 0s, s for seconds and A for meridiem. It uses the correct format, but it gives me the wrong time. My local time is 4:43 and it prints out 4:12:
<?php echo "<p>order processed on " . date("F d, Y g:sA") . "</p>"; ?> Why is it 30 minutes behind?
Hi, I need code that reads from the roles database and then selects which file from these 3 which I want. For example, the user.php file would be loaded if the user has UName = user, Pass = 124, and Roles = User added to the database. But the admin.php and boss.php files would not appear to him.
<?php
session_start();
if(!(isset($_SESSION['User'])))
{
header("Location: index.php");
exit(0);
}
?>
<!DOCTYPE html>
<html>
<body>
<?php
include "config.php";
?>
<!--show for User-->
<?php include 'user.php';?>
<!--show for Admin-->
<?php include 'admin.php';?>
<!--show for Boss-->
<?php include 'boss.php';?>
</body>
</html>
I have a log system that allows 10 logs on each side(Left and right). I am trying to make it so that the left side has the 10 most recent logs, then the right as the next 10. Any ideas? So I need to echo a row from my database with php, but where i need to echo is already inside an echo. This is my part of my code: $con = mysql_connect("$host","$username","$password"); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("main", $con); $result = mysql_query("SELECT * FROM Vendor"); while($row = mysql_fetch_array($result)) { //I need to echo right here .................. but I get a blank page when I try this. Please Help. echo '<option value=$row['vendor_id']>'; echo $row['vendor_id']; echo '</option>'; } mysql_close($con); Result: A Blank page. Thanks in advance! Hi I have a list of states using the array method in a form. The drop down menu works fine. I want to save the user choice,if the form is re-displayed due to a blank field or pattern mismatch. I know I can use the selected=selected, but don't know wher to put the statement: My array is: state_province = array ("list of states", "provinces") Var in my labels array is "state"=>"state" Here is my code for the select/option statement: { if($field == "state") { echo "<div class='province_state'><label for='state' size='10'>* Province/State</label><select>"; foreach($state_province as $state) { echo "\n<option value='$state_province' /> "; echo $state ; echo "</option>"; } echo "</select></div>\n"; } ?Is this the correct code to add and where would I add it? if(@$_POST['state'] == $value) { echo "selected='selected' "; } Hi
I try to echo out random lines of a html file and want after submit password to whole content of the same html file. I have two Problems.
1st Problem When I echo out the random lines of the html file I don't get just the text but the code of the html file as well. I don't want that. I just want the text. How to do that?
for($x = 1;$x<=40;$x++) {
$lines = file("$filename.html");
echo $lines[rand(0, count($lines)-1)]."<br>";
}
I tried instead of "file("$filename.html");" "readfile("$filename.html");" But then I get the random lines plus the whole content. Is there anything else I can use instead of file so that I get the random lines of text without the html code?P.S file_get_contents doesn't work either have tried that one.
2nd Problem:
As you could see in my first problem I have a file called $filename.html. After I submit the value of a password I want the whole content. But it is like the program did forget what $filename.html is. How can I make the program remember what $filename.html is? Or with other words how to get the whole content of the html file?
My code:
if($_POST['submitPasswordIT']){
if ($_POST['passIT']== $password ){
$my_file = file_get_contents("$filename.html");
echo $my_file;
}
else{
echo "You entered wrong password";
}
}
If the password isn't correct I get: You entered wrong password. If the password is correct I get nothing. I probably need to create a path to the file "$filename.html", but I don't know exactly how to do that.
Hi, I am not a PHP programmer. I took on a new client with a simple PHP site, without any databases. The site is up and running on the web. I would like to get it running on my local machine for further development. I have latest version of WAMP installed, running Apache version 2.2.11 and PHP version 5.3.0 I created a directory in the WAMP "www" project directory and it shows up there like it's supposed to when I browse to "localhost" Problem: The home page of website displays text but no, images, styles, footer, header, nav links, etc. Here is the code for the home page: <? define("NAV","home"); require_once('local/local.php'); ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <title>TITLE</title> <meta name="keywords" content=""> <meta name="Description" content=""> <? include("common/dochead.php"); ?> </head> <body onLoad="<? include('common/preloads.php'); ?>"> <!-- ============================ main ============================= --> <div id="main-frame"><div id="main" class="noCollapse"> <? include("common/sign.php"); ?> <div id="right-frame"> <? include("common/navigation.php"); ?> <div id="content-frame"> <div id="content"> <h1>Welcome</h1> <p>This is the content area. This is the content area. This is the content area. </p> </div><!-- end content --> </div><!-- end content-frame --> </div><!-- end right-frame --> <div class="clearFloats"></div> </div><!-- end main --></div><!-- end main-frame --> <? include("common/footer.php"); ?> </body> </html> Any help would be greatly appreciated. I have spent many hours on this. Regards Below is my output on the browser: Student: Kevin Smith (u0867587) Course: INFO101 - Bsc Information Communication Technology Course Mark 70 Grade Year: 3 Module: CHI2550 - Modern Database Applications Module Mark: 41 Mark Percentage: 68 Grade: B Session: AAB Session Mark: 72 Session Weight Contribution 20% Session: AAE Session Mark: 67 Session Weight Contribution 40% Module: CHI2513 - Systems Strategy Module Mark: 31 Mark Percentage: 62 Grade: B Session: AAD Session Mark: 61 Session Weight Contribution 50% Now where it says course mark above it says 70. This is incorrect as it should be 65 (The average between the module marks percentage should be 65 in the example above) but for some stange reason I can get the answer 65. I have a variable called $courseMark and that does the calculation. Now if the $courseMark is echo outside the where loop, then it will equal 65 but if it is put in while loop where I want the variable to be displayed, then it adds up to 70. Why does it do this. Below is the code: Code: [Select] $sessionMark = 0; $sessionWeight = 0; $courseMark = 0; $output = ""; $studentId = false; $courseId = false; $moduleId = false; while ($row = mysql_fetch_array($result)) { $sessionMark += round($row['Mark'] / 100 * $row['SessionWeight']); $sessionWeight += ($row['SessionWeight']); $courseMark = ($sessionMark / $sessionWeight * 100); if($studentId != $row['StudentUsername']) { //Student has changed $studentId = $row['StudentUsername']; $output .= "<p><strong>Student:</strong> {$row['StudentForename']} {$row['StudentSurname']} ({$row['StudentUsername']})\n"; } if($courseId != $row['CourseId']) { //Course has changed $courseId = $row['CourseId']; $output .= "<br><strong>Course:</strong> {$row['CourseId']} - {$row['CourseName']} <strong>Course Mark</strong>" round($courseMark) "<strong>Grade</strong> <br><strong>Year:</strong> {$row['Year']}</p>\n"; } if($moduleId != $row['ModuleId']) { //Module has changed if(isset($sessionsAry)) //Don't run function for first record { //Get output for last module and sessions $output .= outputModule($moduleId, $moduleName, $sessionsAry); } //Reset sessions data array and Set values for new module $sessionsAry = array(); $moduleId = $row['ModuleId']; $moduleName = $row['ModuleName']; } //Add session data to array for current module $sessionsAry[] = array('SessionId'=>$row['SessionId'], 'Mark'=>$row['Mark'], 'SessionWeight'=>$row['SessionWeight']); } //Get output for last module $output .= outputModule($moduleId, $moduleName, $sessionsAry); //Display the output echo $output; I think the problem is that it is outputting the answer of the calculation only for the first session mark. How in the while loop can I do it so it doesn't display it for the first mark only but for all the session marks so that it ends up showing the correct answer 65 and not 72? What is the correct way to write an if/else statement within an echo? I need to alter this so that I can query to see what data is found and if not correct not to echo the rest of the statement. Code: [Select] echo '<td class="productbox"> <h1>' . $product_title . '</h1> </td>'; So from the above code which is echoed within the single quotes, what is the correct way to include an if else check on a value from the database. I know how its done, but just want to save time and write it the correct way within this. Code: [Select] <?php $statisctis=("SELECT date_liked , COUNT(site_id) AS num_site_id FROM `liked` WHERE date_liked < CURDATE() AND date_liked > CURDATE() - INTERVAL 1 WEEK AND site_id='45' GROUP BY date_liked"); $result1 = mysql_query ($statisctis) or die(mysql_error()); ?> in php admin i get result like this date_liked num_site_id 2011-11-28 10:00:29 1 2011-11-29 05:03:17 1 2011-11-30 04:51:37 1 how to echo this in my page? Here i get a COUNT number only Code: [Select] <?php $sum_datas = mysql_result($result1, 0); echo $sum_datas['date_liked']; ?> For some reason, I can only get 1 row to echo out when there are actually multiple rows in the database that should be echoing. Here is what I have: $result = mysql_query("SELECT * FROM boards WHERE boardname='$board' ORDER BY id LIMIT 10"); $post = mysql_fetch_assoc($result); echo stripslashes($post['message']) . "<br>\n" . " --- ".stripslashes($post['username']). " on ".stripslashes($post['time']) ."\n<hr width=90%>\n"; It pulls the one record great, but it only shows one record... I want to keep it to 10 records, thus the LIMIT in there (which I think I did right...), but it won't even show the ones in there right now (under 10, so that's not an issue yet). Hi I once found out how to do this 4 years ago but I lost my backups and memory on the matter. I want this xml to be echoed? out as it is. When I try the standard echo and print, I get the first two lines of the code in gray when I view the page's source. I get no xml at all. Code: [Select] <?xml version="1.0" encoding="utf-8"?> <!DOCTYPE blah [ <!ELEMENT cart (title, items)> <!ELEMENT title (#PCDATA)> <!ELEMENT items (item)+> <!ELEMENT item (prive, deprive, onprive+)> <!ELEMENT security (#PCDATA)> <!ELEMENT answer (#PCDATA)> <!ATTLIST answer correct (yeah) #IMPLIED> ]> this is followed by the standard displaying of xml items could anyone help me over here? thanks in advance... |
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