PHP - How To Display My Images Along With Mysql Database Table
Hi, here's my problem: I am trying to make a simple online buying website and I want to display a table with all the fields for each item. So I got that part down which is to just use mysql_fetch_assoc("SELECT * FROM myTable") and use the html table tags stuff, but now I want to display my images in the table, so here's my code to display my mysql database table in html's table tag along w/ php:
<html> <head> <title>My Online buying website project</title> </head> <body> <?php mysql_connect("localhost","root"); mysql_select_db("myTable"); $imagesArray=array("Apple_iPhone3GS.jpg","Apple_iPhone4.jpg","product3.jpg","product4.jpg","product5.jpg"); $result=mysql_query("SELECT Name, Manufacturer, Price, Description, SimSupport FROM myTable"); if(mysql_num_rows($result))//if there is at least one entry in bellProducts, make a table { print "<table border='border'>"; print "<tr> <th>Name</th> <th>Manufacturer</th> <th>Price</th> <th>Description</th> <th>SimSupport</th> </tr>"; //NB: now output each row of records while($row=mysql_fetch_assoc($result)) { extract($row); print "<tr> <td>$Name</td> <td>$Manufacturer</td> <td>$Price</td> <td>$Description</td><td>$SimSupport</td> </tr>"; }//END WHILE }//END IF ?> </table> </body> </html> *So how do I go about adding my images in this table? Similar Tutorialsi have 8 division (div), i want to display 4 rows in 4 division and the remain 4 rows in the next 4 division here is my code structure for carousel
<div class="nyie-outer"> second row third row
fourth row fifth row sixth row seven throw eighth row
</div><!--/.second four rows here-->
sql code
CREATE TABLE product( php code
<?php how can i echo that result in those rows
Hi guys its me again, I am having a problem that I cant figure out... Here is my code: <?php $sqlCommand = "SELECT image FROM background"; $query = mysqli_query($myConnection, $sqlCommand) or die (mysqli_error()); $sqlCommand2 = "SELECT backgroundimage FROM site"; $query2 = mysqli_query($myConnection, $sqlCommand2) or die (mysqli_error()); while ($row = mysqli_fetch_array($query)) { while ($row2 = mysqli_fetch_array($query2)) { if($row['image'] == $row2['backgroundimage']){ echo '<img src="site_background/'.$row['image'].'" width="75px" height="75px" style="border:2px solid red;" /><br /><br />'; } if($row['image'] != $row2['backgroundimage']){ echo '<img src="site_background/'.$row['image'].'" width="50px" height="50px" style="border:2px solid black;" />'; } } } mysqli_free_result($query); mysqli_free_result($query2); ?> What is will do is get the images from the "backgrounds" table and the image from the "site" table (the current image). I am then wanting to pick out the current image and give it a red border and then display the other left over images smaller with a black border. I can get the images to all display with the black border or the current image to display with a red border but the other images dont show... I have tried mixing things around but I have not been able to get all the images to display with the formatting I want. I dont know if it is a simple syntax error or I am doing things completely wrong... I have been looking at it for so long its just become one big mess of code to me lol Any help to get this working as I want would be great! Cheers Ben I want to display 3 clickable images in a single row which repeats as long as there is data in the database, so far it is displaying a single clickable image from the database. below is all the code.. <table width="362" border="0"> <?php $sql=mysql_query("select * from `publication` GROUP BY `catsue`") or die(mysql_error()); $num=mysql_num_rows($sql); while($rowfor=mysql_fetch_array($sql)) { $cat=$rowfor['catsue']; $pic=mysql_query("select * from `category` where `catsue`='$cat'") or die(mysql_error()); $picP=mysql_fetch_array($pic); $base=basename($picP['title']); ?> <tr> <td width="352" height="88"><table width="408" border="0"> <tr> <td width="113" rowspan="5"><a href="archive_detail.php?id=<?php echo $rowfor['id'];?>&category=<?php echo $rowfor['catsue'];?>"><img src="ad/pic/<?php echo $base;?>" width="100" height="100" border="0"/></a></td> <td width="94">Title</td> <td width="179" height="1"><?php echo $rowfor['catsue'];?> </td> </tr> <tr> <td> </td> <td width="179" height="3"> </td> </tr> <tr> <td> </td> <td width="179" height="8"> </td> </tr> <tr> <td> </td> <td width="179" height="17"> </td> </tr> <tr> <td> </td> <td width="179" height="36"> </td> </tr> </table></td> </tr> <?php }?> </table> Hello everyone, I need help figuring out why im not getting images displayed. I have tried everything I could think of and narrowed down the issue. What Im doing id querying urls from DB for my images. Then using md5 to encrypting the url them before display on web page. Here is the code: Code: [Select] $img = @ md5(mysql_result(mysql_query("SELECT `url` FROM `pictures` WHERE `mls`='{$row['mls']}' ORDER BY `id` LIMIT 1",$avenu->link),0)); When html is displayed I get the hash.jpg. The image failed to be loaded. I want to make a "hint" feature for an app ive been working on. Its a guessing game, and I want to be able to give out hints. Ok so say there is an answer in the database. Say the value is "firetruck". I want to retrieve that value, and obscure it, displaying only one or two letters that are in the word. Or, for a number, say 178, I want to display only one digit from it. Is this possible? Im sure it is, anything seems possible with PHP & Mysql. If so, how could I implement this? Trust me, I read the manuals. The manuals for both PHP and Mysql are so vast and a little advanced for my level.. then again im not complete novice and know the basics plus more of both. Ok the pagination part is all working fine. but i thought id be able to create a heap variables inside the loop then display the images in a table. the only trouble is all variables are grabbing the same img. i need them to grab the 10 different records. thanks Code: [Select] $sql = "SELECT * FROM mongrels_db.gallery ORDER BY id DESC LIMIT $offset, $rowsperpage "; $result = mysql_query($sql, $conn) or trigger_error("SQL", E_USER_ERROR); // while there are rows to be fetched... while ($list = mysql_fetch_array($result)) { $img1=$list['img']; $img2=$list['img']; $img3=$list['img']; $img4=$list['img']; $img5=$list['img']; $img6=$list['img']; $img7=$list['img']; $img8=$list['img']; $img9=$list['img']; $img10=$list['img']; // echo data } // end while echo "<table><tr>"; echo "<td>"."<img src='../gallery/".$img1 ."' width='100' height='100''> "."</td>"; echo "<td>"."<img src='../gallery/".$img2 ."' width='100' height='100''> "."</td>"; echo "<td>"."<img src='../gallery/".$img3 ."' width='100' height='100''> "."</td>"; echo "<td>"."<img src='../gallery/".$img4 ."' width='100' height='100''> "."</td></tr>"; echo "<tr><td>"."<img src='../gallery/".$img5 ."' width='100' height='100''> "."</td>"; echo "<td>"."<img src='../gallery/".$img6 ."' width='100' height='100''> "."</td>"; echo "<td>"."<img src='../gallery/".$img7 ."' width='100' height='100''> "."</td>"; echo "<td>"."<img src='../gallery/".$img8 ."' width='100' height='100''> "."</td>"; echo "<tr><td>"."<img src='../gallery/".$img9 ."' width='100' height='100''> "."</td>"; echo "<td>"."<img src='../gallery/".$img10 ."' width='100' height='100''> "."</td></tr>"; im making a game and i need to show a users money but i dont know how help? Hi there. I have this simple code which displays 5 results. How can i grab each element separately instead of displaying all the results at once. Thanks:) Code: [Select] <?php $query = mysql_query("SELECT product_name, product_price FROM products WHERE product_type = 'laptop' LIMIT 5"); $numrows = mysql_num_rows($query); if ($numrows != 0) { while ($row = mysql_fetch_assoc($query)) { $product_name = $row['product_name']; $product_price = $row['product_price']; echo $product_name . '<br />'; echo $product_price . '<br /><br />'; } } ?> I have got connection to the the mysql database, how do I get the data from the database to display on the webpage Image name will come from database. It can be displayed in divs or table cells. Which is better using divs or table cells? Can anyone post sample code? http://img28.imageshack.us/i/44233003.png/ Basically i have a folder with 100+ images they are NOT all the same extension, what im wanting to do is use PHP to find all the images and put them all in a database. how would i go about doing this? thanks Say a user puts in a support request, and for every request it generates a unqiue string, and enters it into the database. Ok, now say there is a text field, when the user enters their unique string and it finds a match, it displays the data along with it. How can I accomplish this? Im kind of new to mysql, but I know basic SQL. Would be great if somebody could point me in the right direction! Thanks Hello, What I am looking to do is have a series of buttons laid out and when the user 'onClick', a table from my MySQL database will load into a scrollable textbox within the same webpage. I figure I need to go about this using AJAX, which unfortunately I have basically no experience in. For example purposes here is what I have going: This php code basically just displays the table we want to tie the button click to. Code: [Select] <?php include("connect.php"); //Query of facebook database $facebook = mysql_query("SELECT * FROM facebook") or die(mysql_error()); //Output results if(!$facebook) { echo "There was an error running the query: " . mysql_error(); } elseif(!mysql_num_rows($facebook)) { echo "No results returned"; } else { $header = false; echo "<table border='1'>\n"; while($row = mysql_fetch_assoc($facebook)) { if(!$header) { echo "<tr>\n"; foreach($row as $header => $value) { echo "<th>{$header}</th>\n"; } echo "</tr>\n"; } echo "<tr>\n"; foreach($row as $value) { echo "<th>{$value}</th>\n"; } echo "</tr>\n"; } echo "</table>\n"; } mysql_close(); ?> I'd really like to make this as easy as possible. After doing research, I figured that just Javascript would not be enough because it is client-side. It would be great if someone could point me in the right direction. Thanks in advance to anyone who replies. how i want to display data from database to look like this : <table width="633" height="224" border="1"> <tr bgcolor="#999900"> <td width="45">Bil</td> <td width="121">Course_name</td> <td width="83">session</td> <td width="83">start_date</td> <td width="83">end_date</td> <td width="83">notes</td> <td width="89">pre-req</td> </tr> <tr bgcolor="#6A7AEA"> <td rowspan="2">1.</td> <td rowspan="2" bgcolor="#6A7AEA">Math</td> <td>1st session </td> <td>1 jan 11 </td> <td>6 jan 11 </td> <td rowspan="2"> </td> <td rowspan="2"><image icon that will link to the oter site> </td> </tr> <tr> <td bgcolor="#6A7AEA">2nd session </td> <td bgcolor="#6A7AEA">8 jan 11 </td> <td bgcolor="#6A7AEA">15 jan 11 </td> </tr> <tr> <td bgcolor="#0066CC">2.</td> <td bgcolor="#0066CC">English</td> <td bgcolor="#0066CC">1st session </td> <td bgcolor="#0066CC">1 feb 11 </td> <td bgcolor="#0066CC">6 feb 11 </td> <td bgcolor="#0066CC"> </td> <td bgcolor="#0066CC"><image icon that will link to the oter site></td> </tr> <tr> <td rowspan="2" bgcolor="#6A7AEA">3.</td> <td rowspan="2" bgcolor="#6A7AEA">Science</td> <td height="29" bgcolor="#6A7AEA">1st session </td> <td bgcolor="#6A7AEA">8 march 11 </td> <td bgcolor="#6A7AEA">15 march 11 </td> <td rowspan="2" bgcolor="#6A7AEA"> </td> <td rowspan="2" bgcolor="#6A7AEA"><image icon that will link to the oter site></td> </tr> <tr> <td bgcolor="#6A7AEA">2nd session</td> <td bgcolor="#6A7AEA">16 march 11 </td> <td bgcolor="#6A7AEA">21 march 11 </td> </tr> </table> ** all the view data is called from database including the icon image thanks... Ok, I got someone to help me fix this but he had no idea what the error was... I have 2 tables, one called points and the other called members. In members i have got: id name In points i have got: id memberid promo I have the following code: Code: [Select] <?php $con = mysql_connect("localhost","slay2day_User","slay2day"); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("slay2day_database",$con); $sqlquery="SELECT Sum(points.promo) AS score, members.name, members.id = points.memberid Order By members.name ASC"; $result=mysql_query($sqlquery,$con); while ($row = mysql_fetch_array($result)) { //get data $id = $row['id']; $name = $row['name']; $score = $row['score']; echo "<b>Name:</b> $name<br />"; echo "<b>Points: </b> $score<br />" ; echo "<b>Rank: </b>"; if ($name == 'Kcroto1'): echo 'The Awesome Leader'; else: if ($points >= '50'): echo 'General'; elseif ($points >= '20'): echo 'Captain!'; elseif ($points >= '10'): echo 'lieutenant'; elseif ($points >= '5'): echo 'Sergeant'; elseif ($points >= '2'): echo 'Corporal'; else: echo 'Recruit'; endif; endif; echo '<br /><br />'; } ?> I am getting the following error when i do the query in mysql: Code: [Select] #1109 - Unknown table 'points' in field list And when i open the webpage i get the following error: Code: [Select] Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in /home/slay2day/public_html/points/members.php on line 18 Please Help me? Hello Guys, I need help on this problem. output: display images for the values that exist in the table: input: the values in a table's column where they're in the format of "0,0" and the image corresponding to it is named 0,0.png this is my attempt $q is retrieving "0,0" however in order for me to retrieve its image i need to add ".png" to it in order to find it in the directory that the image is located. It tried without ".png" and it also doesn't work. $q = 'SELECT ID FROM table'; $dirname = "directory/"; $images = glob($dirname.$q.".png"); foreach($images as $image) { echo '<img src="'.$image.'" /><br />';
For the last few days I have been trying to work out why my code does not update the MySQL database table. Having tried several variation I have the below but cannot see anything wrong. The rest of the program produces the correct results (displaying what is currently on the table) showing it is connecting to the correct table but altering the table is not working. Any help greatly appreciated. The few lines in question a Code: [Select] // Update the profile data in the database if (!$error) { if (!empty($name)&& !empty($phone) && !empty($address1) && !empty($address2)) { // Only set the picture column if there is a new picture if (!empty($new_picture)) { //if (!empty($postcode)){ $query = "UPDATE antique SET name = '$name', phone = '$phone', address1 = '$address1', address2 = '$address2', postcode = '$postcode', " . " email = '$email', webadd = '$webadd', picture = '$new_picture', username = '" . $_SESSION['username'] . "' WHERE name = '" . $row['name'] ."'"; }} else { $query = "UPDATE antique SET name = '$name', phone = '$phone', address1 = '$address1', address2 = '$address2', postcode = '$postcode', " . " email = '$email', webadd = '$webadd', username = '" . $_SESSION['username'] . "' WHERE name = '" . $row['name'] ."'"; } mysqli_query($dbc, $query) or die("<br>Query $query<br>Failed with error: " . mysqli_error($dbc) . '<br>On line: ' . __LINE__); The whole program is below Code: [Select] <?php error_reporting(E_ALL); session_start(); ?> <?php require_once('appvars.php'); require_once('connectvars1.php'); // Connect to the database $dbc = mysqli_connect(DB_Host, DB_User, DB_Password, DB_Name); if (!isset($_GET['user_id'])) { $query = "SELECT * FROM antique WHERE user_id = '" . $_SESSION['user_id'] . "'"; } else { $query = "SELECT * FROM antique WHERE user_id = '" . $_GET['user_id'] . "'"; } $data = mysqli_query($dbc, $query); if (mysqli_num_rows($data) == 1) { // The user row was found so display the user data $row = mysqli_fetch_array($data); echo '<table>'; if (!empty($row['name'])) { echo '<tr><td class="label">Name:</td><td>' . $row['name'] . '</td></tr>'; } if (!empty($row['phone'])) { echo '<tr><td class="label">Phone:</td><td>' . $row['phone'] . ' </td></tr>'; } if (!empty($row['address1'])) { echo '<tr><td class="label">Address1:</td><td>' . $row['address1'] . ' </td></tr>'; } if (!empty($row['address2'])) { echo '<tr><td class="label">Address2:</td><td>' . $row['address2'] . ' </td></tr>'; } if (!empty($row['postcode'])) { echo '<tr><td class="label">Postcode:</td><td>' . $row['postcode'] . ' </td></tr>'; } if (!empty($row['webadd'])) { echo '<tr><td class="label">Web address:</td><td>' . $row['webadd'] . ' </td></tr>'; } if (!empty($row['email'])) { echo '<tr><td class="label">Email:</td><td>' . $row['email'] . ' </td></tr>'; } if (!empty($row['username'])) { echo '<tr><td class="label">Username:</td><td>' . $row['username'] . ' </td></tr>'; } if (!empty($row['user_id'])) { echo '<tr><td class="label">User ID:</td><td>' . $row['user_id'] . ' </td></tr>'; } echo '</table>'; //echo '<class = "label">USER ID: ' . $_SESSION['user_id'] . ''; if (!isset($_GET['postcode']) || ($_SESSION['postcode'] == $_GET['postcode'])) { echo '<p>Would you like to <a href="index5.php">Go to Homepage</a>?</p>'; } } // End of check for a single row of user results else { echo '<p class="error">There was a bit of a problem accessing your profile.</p>'; } ?> <hr> <?php require_once('appvars.php'); require_once('connectvars1.php'); // Make sure the user is logged in before going any further. if (!isset($_SESSION['user_id'])) { echo '<p class="login">Please <a href="login1.php">log in</a> to access this page.</p>'; exit(); } // Connect to the database $dbc = mysqli_connect(DB_Host, DB_User, DB_Password, DB_Name); if (isset($_POST['submit'])) { // Grab the profile data from the POST $name = mysqli_real_escape_string($dbc, trim($_POST['name'])); $phone = mysqli_real_escape_string($dbc, trim($_POST['phone'])); $address1 = mysqli_real_escape_string($dbc, trim($_POST['address1'])); $address2 = mysqli_real_escape_string($dbc, trim($_POST['address2'])); $postcode = mysqli_real_escape_string($dbc, trim($_POST['postcode'])); $webadd = mysqli_real_escape_string($dbc, trim($_POST['webadd'])); $email = mysqli_real_escape_string($dbc, trim($_POST['email'])); $old_picture = mysqli_real_escape_string($dbc, trim($_POST['old_picture'])); $new_picture = mysqli_real_escape_string($dbc, trim($_FILES['new_picture']['name'])); $new_picture_type = $_FILES['new_picture']['type']; $new_picture_size = $_FILES['new_picture']['size']; $username = mysqli_real_escape_string($dbc, trim($_POST['username'])); $user_id = mysqli_real_escape_string($dbc, trim($_POST['user_id'])); if (!empty($_FILES['new_picture']['tmp_name'])) {list($new_picture_width, $new_picture_height) = getimagesize($_FILES['new_picture']['tmp_name']); } //list($new_picture_width, $new_picture_height) = getimagesize($_FILES['new_picture']['tmp_name']); $error = false; // Validate and move the uploaded picture file, if necessary if (!empty($new_picture)) { if ((($new_picture_type == 'image/gif') || ($new_picture_type == 'image/jpeg') || ($new_picture_type == 'image/pjpeg') || ($new_picture_type == 'image/png')) && ($new_picture_size > 0) && ($new_picture_size <= MM_MAXFILESIZE) && ($new_picture_width <= MM_MAXIMGWIDTH) && ($new_picture_height <= MM_MAXIMGHEIGHT)) { if ($_FILES['new_picture']['error'] == 0) { // Move the file to the target upload folder $target = MM_UPLOADPATH . basename($new_picture); if (move_uploaded_file($_FILES['new_picture']['tmp_name'], $target)) { // The new picture file move was successful, now make sure any old picture is deleted if (!empty($old_picture) && ($old_picture != $new_picture)) { } } else { // The new picture file move failed, so delete the temporary file and set the error flag @unlink($_FILES['new_picture']['tmp_name']); $error = true; echo '<p class="error">Sorry, there was a problem uploading your picture.</p>'; } } } else { // The new picture file is not valid, so delete the temporary file and set the error flag @unlink($_FILES['new_picture']['tmp_name']); $error = true; echo '<p class="error">Your picture must be a GIF, JPEG, or PNG image file no greater than ' . (MM_MAXFILESIZE / 1024) . ' KB and ' . MM_MAXIMGWIDTH . 'x' . MM_MAXIMGHEIGHT . ' pixels in size.</p>'; } } $error = false; // Update the profile data in the database if (!$error) { if (!empty($name)&& !empty($phone) && !empty($address1) && !empty($address2)) { // Only set the picture column if there is a new picture if (!empty($new_picture)) { //if (!empty($postcode)){ $query = "UPDATE antique SET name = '$name', phone = '$phone', address1 = '$address1', address2 = '$address2', postcode = '$postcode', " . " email = '$email', webadd = '$webadd', picture = '$new_picture', username = '" . $_SESSION['username'] . "' WHERE name = '" . $row['name'] ."'"; }} else { $query = "UPDATE antique SET name = '$name', phone = '$phone', address1 = '$address1', address2 = '$address2', postcode = '$postcode', " . " email = '$email', webadd = '$webadd', username = '" . $_SESSION['username'] . "' WHERE name = '" . $row['name'] ."'"; } mysqli_query($dbc, $query) or die("<br>Query $query<br>Failed with error: " . mysqli_error($dbc) . '<br>On line: ' . __LINE__); // Confirm success with the user echo '<p>Your profile has been successfully updated. Would you like to <a href="viewprofile4.php">view your profile</a>?</p>'; mysqli_close($dbc); exit(); } else { echo '<p class="error">You must enter all of the profile data (the picture is optional).</p>'; } } // End of check for form submission else { // Grab the profile data from the database $query="SELECT * FROM antique WHERE user_id= '" . $row['user_id'] . "'"; $data = mysqli_query($dbc, $query); $row = mysqli_fetch_array($data); if ($row != NULL) { $name = $row['name']; $phone = $row['phone']; $address1 = $row['address1']; $address2 = $row['address2']; $postcode = $row['postcode']; $email = $row['email']; $webadd = $row['webadd']; $old_picture = $row['picture']; $username = $_SESSION['username']; $user_id = $row['user_id']; } else { echo '<p class="error">There was a problem accessing your profile.</p>'; } } mysqli_close($dbc); ?> <form enctype="multipart/form-data" method="post" action="<?php echo $_SERVER['PHP_SELF']; ?>"> <input type="hidden" name="MAX_FILE_SIZE" value="<?php echo MM_MAXFILESIZE; ?>" /> <fieldset> <legend>Personal Information</legend> <label for="name">Name:</label> <input type="text" id="name" name="name" value="<?php if (!empty($name)) echo $name; ?>" /><br /> <label for="phone">Phone:</label> <input type="text" id="phone" name="phone" value="<?php if (!empty($phone)) echo $phone; ?>" /><br /> <label for="address1">Address1:</label> <input type="text" id="address1" name="address1" value="<?php if (!empty($address1)) echo $address1; ?>" /><br /> <label for="address2">Address2:</label> <input type="text" id="address2" name="address2" value="<?php if (!empty($address2)) echo $address2; ?>" /><br /> <label for="postcode">Postcode:</label> <input type="text" id="postcode" name="postcode" value="<?php if (!empty($postcode)) echo $postcode; ?>" /><br /> <label for="email">Email:</label> <input type="text" id="email" name="email" value="<?php if (!empty($email)) { echo $email; } else { echo 'No email entered';} ?>" /><br /> <label for="webadd">Web address:</label> <input type="text" id="webadd" name="webadd" value="<?php if (!empty($webadd)) { echo $webadd; } else { echo 'No web address entered';} ?>" /><br /> <input type="hidden" name="old_picture" value="<?php if (!empty($old_picture)) echo $old_picture; ?>" /> <label for="new_picture">Pictu </label> <input type="file" id="new_picture" name="new_picture" /> <?php if (!empty($old_picture)) { echo '<img class="profile" src="' . MM_UPLOADPATH . $old_picture . '" alt="Profile Picture" style: height=100px;" />'; } ?> <br /> <label for="username">Username:</label> <input type="text" id="username" name="username" value="<?php if (!empty($username)) echo $username; ?>" /><br /> <label for="user_id">User ID:</label> <input type="text" id="user_id" name="user_id" value="<?php echo '' . $row['user_id'] . '' ; ?>" /><br /> </fieldset> <input type="submit" value="Save Profile" name="submit" /> </form> <?php echo('<p class="login">You are logged in as ' . $_SESSION['username'] . '. <a href="logout3.php">Log out</a>.</p>'); echo '<class = "label">USER ID: ' . $row['user_id'] . ''; ?> <p><a href="index.php">Return to homepage</a></p> <?php require_once('footer.php'); ?> </body> </html> I receive files/images and loop through an array as follows:
$files=rearrfiles( $_FILES['image']); foreach($files as $key=>$item) { if(is_array($item) && !empty($item['name']) && !empty($item['tmp_name'])){ //$_POST['imageFile']=$item['name']; //echo $item['name']; if($error!=true && !uploadImageB(array('image'=>'image'), DB_EXTENTION."mod_projects_images", $naming, $search = $editId, $dir, 'photo')) { $error=true; $action='Edit'; $message.='<br>Images were unable to be uploaded.'; } } }The function uploadImageB() is as follows: function uploadImageB($info, $table, $naming, $search, $path, $add='uploaded', $maxw=1280, $maxh=1280, $dynamic=true, $id='Id', $thumbMaxWidth = 120) { global $connection; if(!$_POST){ global $HTTP_POST_VARS; @$_POST=$HTTP_POST_VARS;} if(!$_FILES){ global $HTTP_POST_FILES; @$_FILES=$HTTP_POST_FILES;} if(is_array($info) && !empty($info) && $table!='' && !empty($path)) { $error=false; $d=$naming; foreach($info as $key=>$val) { $names[$val]=$dynamic==true?$add.'_'.$result[$id].'_'.$d.'_.'.strtolower(mygetext($_FILES[$val]['name'])):$add.strtolower(mygetext($_FILES[$val]['name'])); if(is_array(@$_FILES[$val]) && !empty($_FILES[$val]['name']) && !empty($names[$val])) { if(!singMove($path, $names[$val], $_FILES[$val], array(), array('gif', 'jpeg', 'jpg', 'png', 'bmp', 'svg'))) { $error=true; break; } else{ $valid[$val]=$names[$val]; resizeImage($path.$names[$val], $maxw, $maxh); make_thumb($path.$names[$val],$names[$val], $path, $thumbMaxWidth); } } } if($error==true) return false; elseif(!empty($valid)) { $connection->info=$info; $connection->input=$valid; $connection->id=array('field'=>$id, 'val'=>addslashes($result[$id])); $connection->makenew($table); $connection->return_db($connection->query); } } return true; }I get the error: Images were unable to be uploaded. Any help? |