jeppers

in a flat text file which i can then pull up the info and use as a voucher code.

hi there i am looking to build a button counter. and i am wondering of you could help. i want the button to +1 every time someone clicks it and also store the value so it can be used as a voucher code. can anyone help thanks

thanks for the help. the problem was the format i had it trying to go in the database like so 2001 20 01 which was wrong when i entered the correct format being 2001-20-01 worked well

please help my head is hurting

<li> <input type="submit" name="submit" value="Send My Posting" /> </li> <?php echo $date; if ($_POST['submit']){ include ('mysql_connect.php'); $sql = "insert into date (dates) values ('$date')"; $result = @mysql_query($sql) or die (mysql_error()); //run query } else { } i am struggling with this error. i have created a dummy database which all i am trying to do is enter the date. which has been selected from pull down menu's i have concatenated them and stored them in a variable call $date. i run the query and i get this as an error message 2010 01 1 Incorrect date value: '2010 01 1' for column 'dates' at row 1 has anyone got any idea on what i am doing wrong. hear is the full code i am using <form action="pulldown.php" method="post"> <label for="date">Date</label> <?php //make the months array $months = array (1 => '01', '02', '03', '04', '05', '06', '07', '08', '09', '10', '11', '12'); //make the days and years array $days = range (1, 31); $years = range (2010, 2012); //make the days pull down menu echo '<select name="days">'; foreach ($days as $key => $value) { echo "<option value=\"$value\">$value</option>\n"; } echo '</select>'; //make that months pull down menus echo '<select name="months">'; foreach ($months as $key => $value) { echo "<option value=\"$value\">$value</option>\n"; } echo '</select>'; //make the year pull down menu echo '<select name="years">'; foreach ($years as $key => $value) { echo "<option value=\"$value\">$value</option>\n"; } echo '</select>'; $days = strtolower($_POST['days']); $months = strtolower($_POST['months']); $years = strtolower($_POST['years']); //create a date variable $date = $years .' '. $months .' '. $days; ?> <li> <input type="submit" name="submit" value="Send My Posting" /> </li> <?php echo $date; if ($_POST['submit']){ include ('mysql_connect.php'); $sql = "insert into date (dates) values ('$date')"; $result = @mysql_query($sql) or die (mysql_error()); //run query } else { } ?>

hi there i would like to ask. does anyone no where i can look for help on extracting the date and time from a pull down menu and adding it to a database. not asking for the answer just a place i could look. hope you can help

well i am working on a photo gallery and it works. pulls info and displays it. thumb nails and main image The problem is when i go on to another page and it displays another 4 images. you click to view these images and instantly the page changes and goes back to the start of the query showing the first images again. what i want to happen is when the image is clicked i want it to stay on that page. i am not sure what to do can you help. please if you need any more info please get in contact. thanks <?php // include MySQL connector function if (! @include('includes/mysql_conn.php')) { echo 'Sorry, page unavailable'; exit; } // define number of columns in table define('COLS', 2); // set maximum number of records per page define('SHOWMAX', 6); // create a connection to MySQL include ('includes/mysql_db.php'); // prepare SQL to get total records $getTotal = 'SELECT COUNT(*) FROM images'; // submit query and store result as $totalPix $total = mysql_query($getTotal); $row = mysql_fetch_row($total); $totalPix = $row[0]; // set the current page $curPage = isset($_GET['curPage']) ? $_GET['curPage'] : 0; // calculate the start row of the subset $startRow = $curPage * SHOWMAX; // prepare SQL to retrieve subset of image details $sql = "SELECT * FROM images LIMIT $startRow,".SHOWMAX; // submit the query (MySQL original) $result = mysql_query($sql) or die(mysql_error()); // extract the first record as an array $row = mysql_fetch_assoc($result); // get the name for the main image if (isset($_GET['image'])) { $mainImage = $_GET['image']; } else { $mainImage = $row['filename']; } // get the dimensions of the main image $imageSize = getimagesize('images/'.$mainImage); ?> <!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=iso-8859-1" /> <title>Salsa In Cumbria <?php if (isset($title)) {echo "&#8212;{$title}";} ?> </title> <link href="includes/journey.css" rel="stylesheet" type="text/css" media="screen" /> </head> <body> <div id="header"> <h1>Salsa Nights</h1> </div> <div id="maincontent"> <h1>Salsa Nights</h1> <div id="wrapper"> <p id="picCount">Displaying <?php echo $startRow+1; if ($startRow+1 < $totalPix) { echo ' to '; if ($startRow+SHOWMAX < $totalPix) { echo $startRow+SHOWMAX; } else { echo $totalPix; } } echo " of $totalPix"; ?></p> <div id="gallery"> <table id="thumbs"> <tr> <!--This row needs to be repeated--> <?php // initialize cell counter outside loop $pos = 0; do { // set caption if thumbnail is same as main image if ($row['filename'] == $mainImage) { $caption = $row['caption']; } ?> <td><a href="<?php echo $_SERVER['PHP_SELF']; ?>?image=<?php echo $row['filename']; ?>"><img src="images/thumbs/<?php echo $row['filename']; ?>" alt="<?php echo $row['caption']; ?>" width="80" height="54" /></a></td> <?php $row = mysql_fetch_assoc($result); // increment counter after next row extracted $pos++; // if at end of row and records remain, insert tags if ($pos%COLS === 0 && is_array($row)) { echo '</tr><tr>'; } } while($row); // end of loop // new loop to fill in final row while ($pos%COLS) { echo '<td> </td>'; $pos++; } ?> </tr> <!-- Navigation link needs to go here --> <tr> <td><?php // create a back link if current page greater than 0 if ($curPage > 0) { echo '<a href="'.$_SERVER['PHP_SELF'].'?curPage='.($curPage-1).'">< Prev</a>'; } // otherwise leave the cell empty else { echo ' '; } ?> </td> <?php // pad the final row with empty cells if more than 2 columns if (COLS-2 > 0) { for ($i = 0; $i < COLS-2; $i++) { echo '<td> </td>'; } } ?> <td> <?php // create a forwards link if more records exist if ($startRow+SHOWMAX < $totalPix) { echo '<a href="'.$_SERVER['PHP_SELF'].'?curPage='.($curPage+1).'">Next ></a>'; } // otherwise leave the cell empty else { echo ' '; } ?> </td> </tr> </table> <div id="main_image"> <p><img src="images/<?php echo $mainImage; ?>" alt="<?php echo $caption; ?>" <?php echo $imageSize[3]; ?> /></p> <p><?php echo $caption; ?></p> </div> </div> </div> </div> </body> </html>

what a fool this is the error $result = @mysql_query ($query) or die(mysql_error()); // Run the query. if (mysql_affected_rows() == 1) { // If it ran OK the query is effection 2 rows not one so all done thanks for all your help today

//chck the form has been submitted if (isset($_POST['submitted'])) { if ($_POST['sure'] == 'yes') { //delete them //make the query $query = "DELETE foods, food_associations FROM foods, food_associations WHERE foods.food_id='$id' AND food_associations.food_id ='$id'"; $result = @mysql_query ($query) or die(mysql_error()); // Run the query. if (mysql_affected_rows() == 1) { // If it ran OK //print the message echo '<h1 "> Delete a food</h1> <p>The food has been deleted.</p><p><br /><br /></p>'; }else{ //if the query did not run ok echo '<h1 ">Sytem error</h1> <p class="error">The food could not be deleted due to a system error.</p>'; // public message echo '<p>' . mysql_error() . '<br /><br />Query: ' . $query . '</p>'; //debugging message. } The code above works and the query is correct. but every time i try to delete it comes up with this message "The food could not be deleted due to a system error" i then check the database and it has been deleted. can u have a look at the code and see if it something very simple thanks

ok ok i was been a fool. thanks for the help

well the script is pulling a $id from another script. // Check for a valid user ID, through GET or POST. if ( (isset($_GET['id'])) && (is_numeric($_GET['id'])) ) { // Accessed through view_users.php $id = $_GET['id']; } elseif ( (isset($_POST['id'])) && (is_numeric($_POST['id'])) ) { // Form has been submitted. $id = $_POST['id']; } else { // No valid ID, kill the script. echo '<h1">Page Error</h1> <p>This page has been accessed in error.</p><p><br /><br /></p>'; exit(); } thanks for spotting the double colon so why do u think there is an empty query

what do u think might be causing my query to be empty. i am not sure what about you. <?php // check the form has been submitted if (isset($_POST['submitted'])) { $errors = array(); // Initialize error array. // Check for a first name. if (empty($_POST['title'])) { $errors[] = 'You forgot to enter the title'; } else { $t =($_POST['title']); } // Check for a last name. if (empty($_POST['description'])) { $errors[] = 'You forgot to enter the description.'; } else { $d =($_POST['description']); } // Check for a last name. if (empty($_POST['price'])) { $errors[] = 'You forgot to enter the price.'; } else { $p =($_POST['price']); } if (empty($errors)) { // If everything's OK. $result = mysql_query($query)or die(mysql_error());; if (mysql_num_rows($result) == 0) { // make query $query = "UPDATE foods SET title='$t', description='$d', price='$p' WHERE food_id=$id"; $result = @mysql_query($query) or die(mysql_error()); if (mysql_affected_rows()==1) {// if ran ok //print message echo '<h1> Edit Food</h1> <p>The food has been updatted</p><p><br /><br /></p>'; } else { //if it did not run ok echo '<h1> System error</h1> <p>Sorry there has been an error and the food could not be updated. sorry</p>'; echo '<p>' . mysql_error() .'<br /><br />Problem:' . $query . '</p>';// debugging message //include ('footer.php') exit(); } }else{ echo '<h1>sorry your food has been updated already</h1>'; } } else { //report the errors echo '<h1>Error!</h1> <p>The follow errors have occurred:<br />'; foreach ($errors as $msg){//print each message echo "-$msg<br />\n"; } echo'</p><p>Please try again.</p><p><br /></p>'; }//end of if }

sorry i am so stuiped i just missed that all out..........

is there any chance you can point me in correct direction, in my code to help me wright that code in so it works.

i am not sure why it won't pick up the id from but it is now starting to get to me so could someone please have a look at this code and maybe tell me where i am going wrong i have 2 pages one where the user can change the oprion in a menu sytem and one where it retives the data using the $id. hear is the code and the user id from the <?php //retrive the users information $query = "SELECT food_id, title, description, price FROM foods WHERE food_id=$id"; $result = @mysql_query($query) or die(mysql_error()); if (mysql_num_rows($result) == 1){//valid id show form //get the users information $row = mysql_fetch_array($result, MYSQL_NUM); hear is the id it is ment to pick up http://localhost//edit_food.php?id=43 so what to do can you help

and my mysql server is 5.0 so i am installing the new version which is 5.1.30 i will get back to if no change

just not sure what you mean i am running apachi mysql. i have opened up the command line for mysql entered mysql --version just comes up with and error. how do u access apachi command line if i need to

well i did the phpifo(); and i got 5.0.45 i then tried finding my mysql version but i just could not find it what do u think i should do. reverting back to my code if that's not the issue. i tried the query in command line and it worked if i put the id number in. so i see the problem being because it does not seem to want to pull the info out as a varable. what do u think the issue could be

i am working on a script where a user can edit information on there menu. this is how it works. they log in and are able to view there menu with a edit button next to every item in there menu. they click and it moved to the edit page. hear is my problem i am trying to gather the info mation from the item they have chosen using the $id as a key so the info can be pulled out. this is my problem i can't seem to get the query correct. could you have a look and see if you can spot what i am missing //retrive the users information <?php $query = "SELECT food_id, title, description, price FROM foods WHERE food_id=$id"; $result = @mysql_query($query) or die(mysql_error()); if (mysql_num_rows($result) == 1){//valid id show form //get the users information $row = mysql_fetch_array($result, MYSQL_NUM); //create the form echo '<h2> Edit Food. <form action="edit_food.php" method="post"> <p>Title: <input type="text" name="title" size="20" maxlength="20" value="'. $row[0] .'" /></p> <p>Description: <input type="text" name="description" size="15" maxlength="30" value="' . $row[1] . '" /></p> <p>Price: <input type="text" name="price" size="20" maxlength="40" value="' . $row[2] . '" /> </p> <p><input type="submit" name="submit" value="Submit" /></p> <input type="hidden" name="submitted" value="TRUE" /> <input type="hidden" name="id" value="' . $id . '" /> </form>'; this is the error message i get 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 '' at line 1 thanks for any help

please can anyone help.

i have 2 tables which need to be updated 1 called food and another food_associations. i want to update the fields in both so when the user views the info they will see there update. sorry i am very tired i will show you the whole script <?php require_once ('mysql_connect.php'); //connect to the database // check the form has been submitted if (isset($_POST['submitted'])) { //check for title if (!empty($_POST['title'])) { $t = ($_POST['title']); } else { $t = FALSE; echo '<p><font color="red">Please enter a food name.</font></p>'; } //check for food description if (!empty($_POST['description'])) { $d =($_POST['description']); } else { $d = FALSE; echo '<p><font color="red">Please enter a description of the food.</font></p>'; } //check for food category if (isset($_POST['types']) && (is_array($_POST['types']))) { $type = TRUE; } else { $type = FALSE; echo '<p><font color="red"> Please select one category.</font></p>'; } //check food price if (!empty($_POST['price'])) { $p = ($_POST['price']); } else { $p = FALSE; echo '<p><font color="red">Please enter a price</font></p>'; } $result = mysql_query($query)or die(mysql_error());; if (mysql_num_rows($result) == 0) { // make query $query = "UPDATE menu SET title='$t', description='$t', price='$p' AS f, food_associations AS fa WHERE f.food_id = fa.food_id AND food_cat_id=$id"; $result = @mysql_query($query) or die(mysql_error()); if (mysql_affected_rows()==1) {// if ran ok //print message echo '<h1> Edit Food</h1> <p>The food has been updatted</p><p><br /><br /></p>'; } else { //if it did not run ok echo '<h1> System error</h1> <p>Sorry there has been a system error and the food could not be updated. sorry</p>'; echo '<p>' . mysql_error() .'<br /><br />Problem:' . $query . '</p>';// debugging message //include ('footer.php') exit(); } } else { //report the errors echo '<h1>Error!</h1> <p>The follow errors have occurred:<br />'; foreach ($errors as $msg){//print each message echo "-$msg<br />\n"; } echo'</p><p>Please try again.</p><p><br /></p>'; }//end of if } //always show the form //retrive the users information $query = "SELECT f.food_id, title, description, price FROM foods AS f, food_associations AS fa WHERE f.food_id = fa.food_id AND fa.food_cat_id=$type"; $result = @mysql_query($query) or die(mysql_error()); if (mysql_num_rows($result) == 1){//valid id show form //get the users information $row = mysql_fetch_array($result, MYSQL_NUM); //create the form echo '<h2> Edit Food. <form action="edit_food.php" method="post"> <p>Title: <input type="text" name="title" size="20" maxlength="20" value="". $row[0] ."" /></p> <p>Description: <input type="text" name="description" size="15" maxlength="30" value="' . $row[1] . '" /></p> <p>Price: <input type="text" name="price" size="20" maxlength="40" value="' . $row[2] . '" /> </p> <p><input type="submit" name="submit" value="Submit" /></p> <input type="hidden" name="submitted" value="TRUE" /> <input type="hidden" name="id" value="' . $id . '" /> </form>'; } else { // Not a valid user ID. echo '<h1>Page Error</h1> <p>This page has been accessed in error.</p><p><br /><br /></p>'; } mysql_close(); // Close the database connection. ?> i want it to display the info that they are tring to change and then once its changed to update the database.

i am not sure if the update query i am using is correct as i am not able to find the error any chance you coul have a look $result = mysql_query($query)or die(mysql_error());; if (mysql_num_rows($result) == 0) { // make query $query = "UPDATE menu SET title='$t', description='$t', price='$p' AS f, food_associations AS fa WHERE f.food_id = fa.food_id AND food_cat_id=$id"; $result = @mysql_query($query) or die(mysql_error()); if (mysql_affected_rows()==1) {// if ran ok //print message echo '<h1> Edit Food</h1> <p>The food has been updatted</p><p><br /><br /></p>'; all help welcome

hi again i am working on the same script but trying to update instead of selecting. my problem is that i have 2x sql code and i am not sure if i am hedding down the correct road to victory. the first set of sql code is trying to update and i feel like i am not in corporating the 2 tables need to be updated. and the second is just trying to select the info to show the user. because there is two sets of code i am just not sure where the issues are: // check the form has been submitted if (isset($_POST['submitted'])) { //check for title if (!empty($_POST['title'])) { $t = ($_POST['title']); } else { $t = FALSE; echo '<p><font color="red">Please enter a food name.</font></p>'; } //check for food description if (!empty($_POST['description'])) { $d =($_POST['description']); } else { $d = FALSE; echo '<p><font color="red">Please enter a description of the food.</font></p>'; } //check for food category if (isset($_POST['types']) && (is_array($_POST['types']))) { $type = TRUE; } else { $type = FALSE; echo '<p><font color="red"> Please select one category.</font></p>'; } //check food price if (!empty($_POST['price'])) { $p = ($_POST['price']); } else { $p = FALSE; echo '<p><font color="red">Please enter a price</font></p>'; } $result = mysql_query($query); if (mysql_num_rows($result) == 0) { // make query $query = "UPDATE title='$t', description='$t', price='$p' AS f, food_associations AS fa WHERE f.food_id = fa.food_id AND food_cat_id=$id"; $result = @mysql_query($query) or die(mysql_error()); if (mysql_affected_rows()==1) {// if ran ok //print message echo '<h1> Edit Food</h1> <p>The food has been updatted</p><p><br /><br /></p>'; } else { //if it did not run ok echo '<h1> System error</h1> <p>Sorry there has been a system error and the food could not be updated. sorry</p>'; echo '<p>' . mysql_error() .'<br /><br />Problem:' . $query . '</p>';// debugging message //include ('footer.php') exit(); } } else { //report the errors echo '<h1>Error!</h1> <p>The follow errors have occurred:<br />'; foreach ($errors as $msg){//print each message echo "-$msg<br />\n"; } echo'</p><p>Please try again.</p><p><br /></p>'; }//end of if } //always show the form //retrive the users information $query = "SELECT f.food_id, title, description, price FROM foods AS f, food_associations AS fa WHERE f.food_id = fa.food_id AND fa.food_cat_id=$type"; $result = @mysql_query($query) or die(mysql_error()); if (mysql_num_rows($result) == 1){//valid id show form //get the users information $row = mysql_fetch_array($result, MYSQL_NUM); please any advice will help thanks

you have to be kidding, it works thanks for the quick responce.

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 'WHERE food_id = fa.food_id AND fa.food_cat_id=1, AND fa.approved = 'Y' ORDER B' at line 2 hope that helps