|
PHP - Where Clause Not Working Probably Simple Answer.
Similar TutorialsHi again all, When i add text to my database through a form, it throws an error if i type 'wouldn't', but not if i type a word without an apostrophe. Is there a simple fix for this please peeps? Many Thanks Me@newbie.com.uk+vat i saw this in a php function page he http://php.net/manual/en/function.stristr.php the line is if(stristr($string, 'earth') === FALSE) { specifically this this code <?php $string = 'Hello World!'; if(stristr($string, 'earth') === FALSE) { echo '"earth" not found in string'; } // outputs: "earth" not found in string ?> what is the significance of the '===' in there does it mean something? is there any difference from the '==' or the '=' ? Hi. I am building an update table form. In this table there are only two fields, Header and Intro. Now I have the form and currently it displays whats already in the database. Now when the user edits these two fields and presses edit. Obviously I want it to update these fields. I have the code here but there is an issue with it. At the end of the update I believe I need a update fields WHERE....... which would be used if there was more than one entrance in the table but there is and only ever will be one entry. WHERE what? I dont have a where anything. Do I need this or can it be omitted somehow? <?php if(isset($_POST['edit'])){ //Process data for validation $header = trim($_POST['header']); $intro = trim($_POST['intro']); //Perform validations next //Prepare data for db insertion $header = mysql_real_escape_string($header); $intro = mysql_real_escape_string($intro); //insert $qUpdateDetails = "UPDATE index SET `header` = '".$header."', `intro` = '".$intro."', WHERE ???????='".?????????."'"; $rUpdateDetails = mysql_query($qUpdateDetails) or die(mysql_error()); //next page $page_edit = "Index page has been edited"; $url = "signed.php?edit='".$page_edit."'" or die(mysql_error()); header("Location: ".$url."") or die(mysql_error()); exit(); } } ?> Hi
I am now on to viewing listings but want to to show listings that are just submitted by that user and display them on their profile page but can't get the WHERE clause working
Below is the coding for what I have
Profile page
echo "<p><a href='view-private-listings.php?id={$_SESSION['user_id']}'>View Listings</a></p>";
Add Listing page HTML
Submitted By: <input type="text" name="submittedby">View Listings Page
<?php
ini_set('display_startup_errors',1);
ini_set('display_errors',1);
error_reporting(-1);
?>
<?php
$title = "Private Seller Listings";
include ( 'includes/header.php' );
?>
<?php
require_once("functions.php");
$con=mysqli_connect("host","username","password","database");
// Check connection
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
// Start a session for error reporting
session_start();
$submittedby=$_POST['submittedby'];
$listingtitle=$_POST['listingtitle'];
$query = mysqli_query($con,"SELECT listingtitle FROM privatelistings WHERE item LIKE '%$submittedby%'");
echo "<p>Title: {$listingtitle['listingtitle']}</p>";
mysqli_close($con);
?>
<?php include( 'includes/footer.php' ); ?>
All I get is the following error
Notice: Undefined index: listingtitle in /zacs-car-site/view-private-listings.php on line 26
Thank you in advance
Kind regards
Ian
hi, how can I add more than one ID to a WHERE clause? Code: [Select] "SELECT * FROM t_mytable WHERE id_product = 1 "; I need to select products with different id's i.e Code: [Select] "SELECT * FROM t_mytable WHERE id_product = 1,2,3,4,5 "; Thanks This may be super simple, but it's stumping me. Is this an If/Then clause: Code: [Select] ($month != 1 ? $month - 1 : 12) Is it saying if $month does not equal 1 then $month-1; if so, then what does the colon mean? If not, then what is this piece of code doing? I am running PHP Version 5.3.1. The following Code does not write to my database. It is code that I took from the PHP Pocket Reference from O'Reilly but it does not work... What's wrong with this code? Thanks in advance. Guy <?php if($vote && !$already_voted) SetCookie('already_voted',1); ?> <html> <head> <title>Name the Baby</title> </head> <h3>Name the Baby</h3> <form action="baby.php" method="POST"> <p>Suggestion: <input type="text" name="new_name"/> </p> <input type="submit" value="Submit idea and/or vote"/> <?php mysql_pconnect("localhost","root","password"); $db = "babynames"; $table = "baby_names"; if($new_name) { if(!mysql_db_query($db, "insert into $table values ('$new_name',0)")) { echo mysql_errno().': '. mysql_error()."<br />\n"; } } if($vote && $already_voted) { echo '<p><b>Hey, you voted already '; echo "Vote ignored.</b></p>\n"; } else if($vote) { if(!mysql_db_query($db, "update $table set votes=votes+1 where name='$vote'")) { echo mysql_errno().': '. mysql_error()."<br />\n"; } } $result=mysql_db_query($db,"select sum(votes) as sum from $table"); if($result) { $sum = (int) mysql_result($result,0,"sum"); mysql_free_result($result); } $result=mysql_db_query($db, "select * from $table order by votes DESC"); echo <<<EOD <table border="0"><tr><th>Vote</th> <th>Idea</th><th colspan="2">Votes</th></tr> EOD; while($row=mysql_fetch_row($result)) { echo <<<FOO <tr><td align="center"> <input type="radio" name="vote" value="$row[0]"></td> <td>$row[0]</td> <td align="right">$row[1]</td> <td> FOO; if ($sum && (int)$row[1]) { $per = (int)(100 * $row[1]/$sum); echo '<img src="bline.gif" height=12 '; echo "width=$per> $per %</td>"; } echo "</tr>\n"; } echo "</table>\n"; mysql_free_result($result); ?> <input type="submit" value="Submit idea and/or vote" /> <input type="reset" /> </form> </body></html> Hey all,
This code was given to me by my client who swears it works, yet I can't seem to get it to function.
<?php
$now = time(); // or your date as well
$your_date = strtotime("2016-06-01");
$datediff = $now - $your_date;
$referrals = number_format(162250 + (floor($datediff/(60*60*24)) * 527) + (87920 + (floor($datediff/(60*60*24)) * 45)));
?>
//javascript:
<script>
//vars from template
var referrals = "<?php echo $referrals;?>";
var currentdate = new Date();
jQuery("document").ready(function() {
jQuery ("#sp-menu > div > nav > ul > li:nth-child(1) > a").html("Home");
jQuery(".referrals").html(referrals);
jQuery(".current_date").html((currentdate.getMonth()+1) + "/" + currentdate.getDate() + "/" + currentdate.getFullYear());
jQuery(window).on("scroll", function() {
var scrollPos = jQuery(window).scrollTop();
if (scrollPos <= 0) {
jQuery(".counter").fadeIn();
jQuery(".sec-nav").fadeIn();
} else {
jQuery(".counter").fadeOut();
jQuery(".sec-nav").fadeOut();
}
});
});
</script>
I've got to be missing something pretty basic here.. considering the query is pretty basic. I'm trying to figure out how to pull a query as an array so I can compare it against another array (array_diff) I'm doing a mysql_fetch_array, and I'm getting an error ( mysql_fetch_array(): supplied argument is not a valid MySQL result resource): Quote $checker = "SELECT ID FROM edible_uses"; $result2 = mysql_fetch_array($checker) or die(mysql_error()); //echoing to see if I'm getting what I need. echo $row['ID']; I've done a mysql_query and I get results. The table name and all that is correct. I'm stumped. This seems like a pretty simple query? I tried mysql_fetch_assoc as well. Same result? I tried it with an extra set of parenthesis around it. nope. Hi, I have a php form that I use to try to get matching data from the database that I put into the form. So if I enter date of birth 9-4-80 and first name Dave and lastname Smith. When I submit it the code should pull all of the matching terms out of the database and display. Now I get the following error when I submit the form. Query: Resource id #2 Failed with error: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'Resource id #2' at line 1 any help Greatly appreciated. thank you! here is the code Code: [Select] <?php ini_set ("display_errors", "1"); error_reporting(E_ALL); $host = " "; $database = " "; $username = " "; $password = " "; $tbl_name = "users"; $conn = mysql_connect($host, $username, $password) or die("Could not connect: " . mysql_error()); if($conn) { mysql_select_db($database); echo "connected to database!!"; } else { echo "failed to select database"; } //include('bouncer.php'); //$currentUser = $_SESSION['myusername']; if(isset($_POST['submit'])) { $first = mysql_real_escape_string( $_POST['first']); $last = mysql_real_escape_string( $_POST['last']); $dob = mysql_real_escape_string( $_POST['dob']); //THE SEARCH FUNCTION $sql = mysql_query ( "SELECT * FROM users WHERE firstname LIKE '%$first%' OR lastname LIKE '%$last%' OR dob LIKE '%$dob%' ") or die(mysql_error()); $result = mysql_query($sql) or die( "<br>Query: $sql<br>Failed with error: " . mysql_error() ); if (!$result) { echo "Could not successfully run query ($sql) from DB: " . mysql_error(); exit; } if (mysql_num_rows($result) == 0) { echo "No rows found, nothing to print so am exiting"; exit; } while ($row = mysql_fetch_assoc($result)) { echo $row["firstname"]; echo $row["lastname"]; echo $row["dob"]; } }//if(isset($_POST['submit'])) ?> <html> <body> <form action="login_success8.php" method="post"> <p> <input type="text" name="first" size="20" /> First name<br /> <input type="text" name="last" size="20" /> Last name<br /> <input name="dob" type="text" size="20" id="dob" /> Date of Birth<br /> <input type="submit" name="submit" value="Search" /> <input type="reset" value="Reset fields" /> </p> </form> </body> </html> Hi, I am new to php, and I have run into a problem. The tutorial I am using has provided me with this exact code. But it does not work for me. Its very simple. Here is the HTML page: Code: [Select] <body> <FORM ACTION="welcome.php" METHOD=POST> First Name: <INPUT TYPE=TEXT NAME="name"> <INPUT TYPE=SUBMIT VALUE="GO"> </FORM> </body> And here is the php page: <body> <?php echo( "Welcome to our Web site, $name!" ); ?> </body> You can see the problem live at <http://www.freewaycreative.com/test> (dont mind the digits below) The name just does not show. Anyone know why? Thanks! The following mysql query is not returning rows like I expect it to. '$update_field' is a variable, matching an actual field name in table 'users'. 'user_task[1]' is an integer value. What am I missing here? Code: [Select] $query_update_user = "UPDATE users SET ".$update_field." = 'Y' WHERE user_no = '".$user_task[1]."'"; I'm trying my first PHP code :
<!DOCTYPE html>
<?php
</body>
Result : nothing, just a blank screen ... Am I missing something ?
Regards, Martin Hi, I have a SQL statement: $sql='SELECT PROPID, ADDRESS_1, ADDRESS_2, TOWN, POSTCODE1, POSTCODE2, BEDROOMS, BATHROOMS, RECEPTIONS, PRICE, TRANS_TYPE_ID FROM properties WHERE TRANS_TYPE_ID=2 AND BEDROOMS =3 AND PROP_SUB_ID IN (1,2,3,4,5,6,21,22,23,24,27,30,95,128,131) ORDER BY DATE_ADDED, PROPID DESC'; I want to be able to get the fields and their corresponding values as such: $result['TRANS_TYPE_ID']=2; $result['BEDROOMS']=3; And then capture the values in PROB_SUB_IN IN(*) How can I do this? Any help will be greatly appreciated. Thanks I am brand new to mysql and php but I have created a database and loaded two tables using cPanel and phpMyAdmin. Now I need some programs to access my data. I have a couple of simple ones that work, but I can't figure out what I really need, I am trying to Select a table Where the Value is a $variable, not a fixed value. Of course the end result will be to pass the value from a Form, but I have to get this to work first. <?php // Connect to database============================= include("connect_db.php"); $table='airplanes'; $amano='123456' $iden='1' // Send query =========================================================== // $result = mysql_query("SELECT * FROM {$table} where ama='123456'"); == this works // $result = mysql_query("SELECT * FROM {$table} where ama='940276'"); == this works // $result = mysql_query("SELECT * FROM {$table} where id='1'"); // this works // $result = mysql_query("SELECT * FROM {$table} where id = '{$iden}'"); == doesnt work // $result = mysql_query("SELECT * FROM {$table} where id = $iden"); == doesnt work // $result = mysql_query("SELECT * FROM {$table} where id = ($iden)"); // == doesnt work // $result = mysql_query("SELECT * FROM {$table} where id = $iden"); // == doesnt work // $result = mysql_query("SELECT * FROM {$table} where ama='$amano'"); // == doesnt work $result = mysql_query("SELECT * FROM {$table} where ama=($amano)"); // == doesnt work Thanks Hi, I have an empty array and i populated the array through code and i printed to check if everything ok. So far so good. I now have an array say $ids = ('123','456'). Now, i have a table in db that has ids as well, I need to make a query to check for id's that are not inside the array, so if the db table have '123','456','789' i want the result of the query to be only the last one '789'. $arr = array(); array_push($arr,$previd2); print_r ($arr); I get the result as follows: Array ( [0] => 533349 [1] => 533355 ) so what query i can use to get the other id's from db table that is not part of that array? here is what i tried: $qry = "select staffname, joindate from staff WHERE OracleID NOT IN ($arr)"; $answ = mysqli_query($con, $qry); I get error: Notice : Array to string conversion in C:\xampp\htdocs\AttendanceSystem\login\reportsforallid.php Edited March 30, 2020 by ramiwahdan more info Code: [Select] $search = "`title` LIKE '%Silvin AND wts%'"; I want to search for both of those keywords why not work? well I lowered my standards massively. LOL. I decided to google, "php upload and display image", instead of "php gallery". Here is the error I get Warning: copy(/images/sheila.jpg) [function.copy]: failed to open stream: No such file or directory in /hermes/bosweb/web173/b1739/sl.brendansite1/public_html/ealike2/smallgallery/smallgallery.php on line 59 and here is the script that I mostly understand. I thought it was the ..images/, but now I don't know what it is. any help greatly appreciated. thank you. below is the code for the page. Code: [Select] <!-- Start PHP Code For Image Upload --> <?php //define a maxim size for the uploaded images in Kb define ("MAX_SIZE","5060"); //This function reads the extension of the file. It is used to determine if the file is an image by checking the extension. function getExtension($str) { $i = strrpos($str,"."); if (!$i) { return ""; } $l = strlen($str) - $i; $ext = substr($str,$i+1,$l); return $ext; } //This variable is used as a flag. The value is initialized with 0 (meaning no error found) //and it will be changed to 1 if an errro occures. //If the error occures the file will not be uploaded. $errors=0; //checks if the form has been submitted if(isset($_POST['Submit'])) { //reads the name of the file the user submitted for uploading $image=$_FILES['image']['name']; //if it is not empty if ($image) { //get the original name of the file from the clients machine $filename = stripslashes($_FILES['image']['name']); //get the extension of the file in a lower case format $extension = getExtension($filename); $extension = strtolower($extension); //if it is not a known extension, we will suppose it is an error and will not upload the file, //otherwise we will do more tests if (($extension != "jpg") && ($extension != "jpeg") && ($extension != "png") && ($extension != "gif")) { //print error message echo '<h2>Unknown extension!</h2>'; $errors=1; } else { //get the size of the image in bytes //$_FILES['image']['tmp_name'] is the temporary filename of the file //in which the uploaded file was stored on the server $size=filesize($_FILES['image']['tmp_name']); //compare the size with the maxim size we defined and print error if bigger if ($size > MAX_SIZE*1024) { echo '<h2>You have exceeded the file size limit! Please reduce the image size to 100 Kb or less!</h2>'; $errors=1; } //we will give an unique name, for example the time in unix time format $image_name=$filename; //the new name will be containing the full path where will be stored (images folder) $newname="../images/".$image_name; //we verify if the image has been uploaded, and print error instead $copied = copy($_FILES['image']['tmp_name'], $newname); if (!$copied) { echo '<h2>Copy unsuccessful!</h2>'; $errors=1; }}}} //If no errors registred, print the success message if(isset($_POST['Submit']) && !$errors) { echo "<h2>File Uploaded Successfully!</h2><br />"; echo "<img src='http://ealike.com/images/<?php echo $image_name; ?> />"; } ?> <!-- End PHP Code For Image Upload --> <!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> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Untitled Document</title> </head> <body> <!-- Start Image Upload Form --> <form name="newad" method="post" enctype="multipart/form-data" action=""> <input type="file" name="image"> <input name="Submit" type="submit" value="Upload image"> </form> <!-- End Image Upload Form --> </body> </html> Hi, I'm trying to make a simple slideshow with forward and back buttons that change the image inside a div in php. When I press the "further" button, it jumps from index[0] to index[1] and never shows the first image. Then when I click more it doesn't go forward. When I click back, it goes back to a black screen div. Any help getting this to work is GREATLY appreciated because I've been trying for 2 days with tutorials and can't get it. Thanks. Derek Here are the php parts that are relevant, my page was too large and confusing to include. first, the $background array where I store my images. Code: [Select] $background = array ( "<img src='sundragon_environments/ocean/ocean1_FRAME.jpg'/>", "<img src='sundragon_environments/ocean/ocean1_FRAME2.jpg'/>" ); then the code to move the images forward and backwards if the buttons are pressed. Code: [Select] if(!empty($_POST['further'])) { $currentBackground=next($background); } elseif(!empty($_POST['back'])) { $currentBackground=prev($background); } and now to echo out the images inside the div. Code: [Select] <div id="background"><?php echo $currentBackground;?></div> I have a form on my website which actions login.php. The login.php code is below: <?php include('includes/classes.php.inc'); session_start(); $link = new BaseClass(); $data = $link->query("SELECT * FROM logins"); $pass_accepted = false; if($_REQUEST['username'] && $_REQUEST['password']){ $username = $_REQUEST['username']; $password = $_REQUEST['password']; while($row = mysql_fetch_array($data)){ if(($row['username']==$useranme)&&($row['password']==$password){ echo 'Password correct!'; $_SESSION['loggedin']=true; $pass_accepted = true; } } } else { echo 'You did not enter a username or password!'; } if(!$pass_accepted){ echo 'Your password is incorrect'; } echo '<br>Please <a href="index.php">click here</a> to return to page'; ?> I have checked that my references are all correct however even when I enter the correct password it returns saying the password is incorrect. Any idea on why this could be? I am happy to answer any follow up questions. Regards |
|||||||