kev wood Posted November 21, 2008 Share Posted November 21, 2008 i am trying to update an entry in a db by the id of the row. i only want to update one of the columns in the table. $sql="UPDATE general SET image= '$consname2' WHERE id= '$id'" or die(mysql_error()); $query = mysql_query($sql)or die(mysql_error()); i am trying to over write the data that is already in the table with a new update for the table. do i need to empty the cell first? Quote Link to comment https://forums.phpfreaks.com/topic/133651-will-not-update/ Share on other sites More sharing options...
Adam Posted November 21, 2008 Share Posted November 21, 2008 What is happening? change this line to: $sql="UPDATE general SET image= '$consname2' WHERE id= '$id'"; Adam Quote Link to comment https://forums.phpfreaks.com/topic/133651-will-not-update/#findComment-695348 Share on other sites More sharing options...
Maq Posted November 21, 2008 Share Posted November 21, 2008 $sql="UPDATE general SET image= '$consname2' WHERE id= '$id'"; mysql_query($sql)or die(mysql_error()); Quote Link to comment https://forums.phpfreaks.com/topic/133651-will-not-update/#findComment-695349 Share on other sites More sharing options...
kev wood Posted November 21, 2008 Author Share Posted November 21, 2008 still getting the same result. no errors and the table is not getting updated. i am posting the id from a previous page with hidden form elements. i am echoing this out at the top of the page so i know it is being sent across. the image is being stored on the server also it is just no updating the existing data in the table. it does not matter that i am only trying to update one coloumn out of the other 5? Quote Link to comment https://forums.phpfreaks.com/topic/133651-will-not-update/#findComment-695362 Share on other sites More sharing options...
Adam Posted November 21, 2008 Share Posted November 21, 2008 No. The "where" determines which records get updated - whether it be 1 or 1000. $query = mysql_query($sql)or die(mysql_error()); Quote Link to comment https://forums.phpfreaks.com/topic/133651-will-not-update/#findComment-695367 Share on other sites More sharing options...
kev wood Posted November 21, 2008 Author Share Posted November 21, 2008 that what i was thinking. so why will this not work. i am telling it to update the image column with id # it displayes the uploaded image but will not update the db? Quote Link to comment https://forums.phpfreaks.com/topic/133651-will-not-update/#findComment-695371 Share on other sites More sharing options...
xangelo Posted November 21, 2008 Share Posted November 21, 2008 Ahh, so the id is a numerical value? Get rid of the single quotes around the id and see if that works. The quotations are meant to define a string value in MySQL. $sql="UPDATE general SET image= '$consname2' WHERE id= $id"; mysql_query($sql)or die(mysql_error()); Quote Link to comment https://forums.phpfreaks.com/topic/133651-will-not-update/#findComment-695377 Share on other sites More sharing options...
kev wood Posted November 21, 2008 Author Share Posted November 21, 2008 still the same result. no errors and no update. this is strange if it not working why no errors. Quote Link to comment https://forums.phpfreaks.com/topic/133651-will-not-update/#findComment-695383 Share on other sites More sharing options...
xangelo Posted November 21, 2008 Share Posted November 21, 2008 Hmm.. are you sure the variables are being passed with the correct values? Quote Link to comment https://forums.phpfreaks.com/topic/133651-will-not-update/#findComment-695389 Share on other sites More sharing options...
Adam Posted November 21, 2008 Share Posted November 21, 2008 Have you changed it to: $query = mysql_query($sql)or die(mysql_error()); (Note $query =) ?? Quote Link to comment https://forums.phpfreaks.com/topic/133651-will-not-update/#findComment-695395 Share on other sites More sharing options...
kev wood Posted November 21, 2008 Author Share Posted November 21, 2008 yes i only have one article in the db at the min, the page before displays the entries with the id number. the id number is inserted into a text box then this submitted to the next page where the same value is echoed out on to the page. i am not getting any connect errors or errors on the update. Quote Link to comment https://forums.phpfreaks.com/topic/133651-will-not-update/#findComment-695400 Share on other sites More sharing options...
kev wood Posted November 21, 2008 Author Share Posted November 21, 2008 the code now looks like this. $sql="UPDATE general SET image= '$consname2' WHERE id= $id"; mysql_query($sql)or die(mysql_error()); $query = mysql_query($sql)or die(mysql_error()); Quote Link to comment https://forums.phpfreaks.com/topic/133651-will-not-update/#findComment-695403 Share on other sites More sharing options...
Maq Posted November 21, 2008 Share Posted November 21, 2008 the code now looks like this. $sql="UPDATE general SET image= '$consname2' WHERE id= $id"; mysql_query($sql)or die(mysql_error()); $query = mysql_query($sql)or die(mysql_error()); Can you please echo $sql; to make sure you have the correct query with valid values. Quote Link to comment https://forums.phpfreaks.com/topic/133651-will-not-update/#findComment-695409 Share on other sites More sharing options...
kev wood Posted November 21, 2008 Author Share Posted November 21, 2008 it has decided to through up an error now which is You have an error in your SQL syntax near '' at line 1 Quote Link to comment https://forums.phpfreaks.com/topic/133651-will-not-update/#findComment-695419 Share on other sites More sharing options...
Maq Posted November 21, 2008 Share Posted November 21, 2008 Please echo your query like I said before... (You have 2 spaces between $consname2 and WHERE if that makes a difference.) $sql="UPDATE general SET image= '$consname2' WHERE id= $id"; echo $sql; mysql_query($sql)or die(mysql_error()); $query = mysql_query($sql)or die(mysql_error()); Quote Link to comment https://forums.phpfreaks.com/topic/133651-will-not-update/#findComment-695424 Share on other sites More sharing options...
xangelo Posted November 21, 2008 Share Posted November 21, 2008 echo 'consname2: '.$consname2.'<br>'; echo 'id: '.$id.'<br>'; $sql='UPDATE general SET image="'.$consname2.'" WHERE id= '.$id; echo $sql; From what it looks like, one of the values aren't being passed properly. What is the output of the above section of code I included? Quote Link to comment https://forums.phpfreaks.com/topic/133651-will-not-update/#findComment-695425 Share on other sites More sharing options...
Adam Posted November 21, 2008 Share Posted November 21, 2008 $sql="UPDATE general SET image= '$consname2' WHERE id= $id"; $query = mysql_query($sql)or die(mysql_error()); echo 'SQL: ' . $sql . '<br />Affected rows: ' . mysql_affected_rows($query); try that .. tell us what it says.. Quote Link to comment https://forums.phpfreaks.com/topic/133651-will-not-update/#findComment-695428 Share on other sites More sharing options...
kev wood Posted November 21, 2008 Author Share Posted November 21, 2008 xangelo: this is what echoed out when i put your section of code in consname2: id: 1 UPDATE general SET image="" WHERE id= 1 the page this is on is the upload page for the image so the image is processed and uploaded and the path is stored into the variable $consname2. the images are uploaded as it is displaying the thumbnail of the image whci it could not do if it was not being uploaded. mac: i have tried to echo out the sql but i have been getting the error near line one since. Quote Link to comment https://forums.phpfreaks.com/topic/133651-will-not-update/#findComment-695437 Share on other sites More sharing options...
kev wood Posted November 21, 2008 Author Share Posted November 21, 2008 MrAdam: changed it to what you said and got this You have an error in your SQL syntax near '' at line 1 Quote Link to comment https://forums.phpfreaks.com/topic/133651-will-not-update/#findComment-695441 Share on other sites More sharing options...
revraz Posted November 21, 2008 Share Posted November 21, 2008 Not sure how you get an error from a echo.. $consname2 is empty. Quote Link to comment https://forums.phpfreaks.com/topic/133651-will-not-update/#findComment-695442 Share on other sites More sharing options...
Adam Posted November 21, 2008 Share Posted November 21, 2008 We already know what we need, as revraz said, $consname2 is empty. Where is $consname2 given a value? Quote Link to comment https://forums.phpfreaks.com/topic/133651-will-not-update/#findComment-695444 Share on other sites More sharing options...
kev wood Posted November 21, 2008 Author Share Posted November 21, 2008 $consname2 is set just before the sql takes place. the images are being stored on the server and $consname2 is set to the path where the image is stored. i can post a link to the page in question if you like Quote Link to comment https://forums.phpfreaks.com/topic/133651-will-not-update/#findComment-695447 Share on other sites More sharing options...
Adam Posted November 21, 2008 Share Posted November 21, 2008 Could you post the code where $consname2 is set? Something isn't going right there! Quote Link to comment https://forums.phpfreaks.com/topic/133651-will-not-update/#findComment-695449 Share on other sites More sharing options...
kev wood Posted November 21, 2008 Author Share Posted November 21, 2008 changed the code to this echo "$consname2"; $sql=("UPDATE general SET image= '$consname2' WHERE id= $id"); $query = mysql_query($sql)or die(mysql_error()); and this is what was echoed out. image/thumb638.jpgYou have an error in your SQL syntax near '' at line 1 Quote Link to comment https://forums.phpfreaks.com/topic/133651-will-not-update/#findComment-695451 Share on other sites More sharing options...
kev wood Posted November 21, 2008 Author Share Posted November 21, 2008 the page is working fine on the insert the only thing i have changed on this page is the sql query and it is not liking the sql it looks like. here is the full page of code <?php $id = $_POST['id']; echo "$id"; set_time_limit(0); $link = mysql_connect(localhost, xxxxxxxx, xxxxxxxx) or die("Could not connect to host."); mysql_select_db(xxxxxxxxdb) or die("Could not find database."); //define a maxim size for the uploaded images define ("MAX_SIZE","500"); // define the width and height for the thumbnail // note that theese dimmensions are considered the maximum dimmension and are not fixed, // because we have to keep the image ratio intact or it will be deformed define ("WIDTH","150"); define ("HEIGHT","120"); // this is the function that will create the thumbnail image from the uploaded image // the resize will be done considering the width and height defined, but without deforming the image function make_thumb($img_name,$filename,$new_w,$new_h) { //get image extension. $ext=getExtension($img_name); //creates the new image using the appropriate function from gd library if(!strcmp("jpg",$ext) || !strcmp("jpeg",$ext)) $src_img=imagecreatefromjpeg($img_name); if(!strcmp("png",$ext)) $src_img=imagecreatefrompng($img_name); if(!strcmp("gif",$ext)) $src_img=imagecreatefromgif($img_name); //gets the dimmensions of the image $old_x=imageSX($src_img); $old_y=imageSY($src_img); // next we will calculate the new dimmensions for the thumbnail image // the next steps will be taken: // 1. calculate the ratio by dividing the old dimmensions with the new ones // 2. if the ratio for the width is higher, the width will remain the one define in WIDTH variable // and the height will be calculated so the image ratio will not change // 3. otherwise we will use the height ratio for the image // as a result, only one of the dimmensions will be from the fixed ones $ratio1=$old_x/$new_w; $ratio2=$old_y/$new_h; if($ratio1>$ratio2) { $thumb_w=$new_w; $thumb_h=$old_y/$ratio1; } else { $thumb_h=$new_h; $thumb_w=$old_x/$ratio2; } // we create a new image with the new dimmensions $dst_img=ImageCreateTrueColor($thumb_w,$thumb_h); // resize the big image to the new created one imagecopyresampled($dst_img,$src_img,0,0,0,0,$thumb_w,$thumb_h,$old_x,$old_y); // output the created image to the file. Now we will have the thumbnail into the file named by $filename if(!strcmp("png",$ext)) imagepng($dst_img,$filename); else imagejpeg($dst_img,$filename); if (!strcmp("gif",$ext)) imagegif($dst_img,$filename); //destroys source and destination images. imagedestroy($dst_img); imagedestroy($src_img); } // This function reads the extension of the file. // It is used to determine if the file is an image by checking the extension. function getExtension($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 error 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['cons_image']['name']; // if it is not empty if ($image) { // get the original name of the file from the clients machine $filename = stripslashes($_FILES['cons_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, print an error message //and will not upload the file, otherwise we continue if (($extension != "jpg") && ($extension != "jpeg") && ($extension != "png") && ($extension != "gif")) { echo '<h1>Unknown extension! Please use .gif, .jpg or .png files only.</h1>'; $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 $size=getimagesize($_FILES['cons_image']['tmp_name']); $sizekb=filesize($_FILES['cons_image']['tmp_name']); //compare the size with the maxim size we defined and print error if bigger if ($sizekb > MAX_SIZE*1024) { echo '<h1>You have exceeded the 1MB size limit!</h1>'; $errors=1; } $rand= rand(0, 1000); //we will give an unique name, for example a random number $image_name=$rand.'.'.$extension; //the new name will be containing the full path where will be stored (images folder) $consname="image/".$image_name; $consname2="image/thumb".$image_name; $copied = copy($_FILES['cons_image']['tmp_name'], $consname); $copied = copy($_FILES['cons_image']['tmp_name'], $consname2); echo "$consname2"; $sql=("UPDATE general SET image= '$consname2' WHERE id= $id"); $query = mysql_query($sql)or die(mysql_error()); echo 'SQL: ' . $sql . '<br />Affected rows: ' . mysql_affected_rows($query); //we verify if the image has been uploaded, and print error instead if (!$copied) { echo '<h1>Copy unsuccessfull!</h1>'; $errors=1; } else { // the new thumbnail image will be placed in images/thumbs/ folder $thumb_name=$consname2 ; // call the function that will create the thumbnail. The function will get as parameters //the image name, the thumbnail name and the width and height desired for the thumbnail $thumb=make_thumb($consname,$thumb_name,WIDTH,HEIGHT); } } } } //If no errors registred, print the success message and show the thumbnail image created if(isset($_POST['Submit']) && !$errors) { echo "<h5>Thumbnail created Successfully!</h5>"; echo '<img src="'.$thumb_name.'">'; echo $id; } echo "<form name=\"newad\" method=\"post\" enctype=\"multipart/form-data\" action=\"\">"; echo "<input type=\"file\" name=\"cons_image\" >"; echo "<input name=\"Submit\" type=\"submit\" id=\"image1\" value=\"Upload image\" />"; echo "</form>"; ?> Quote Link to comment https://forums.phpfreaks.com/topic/133651-will-not-update/#findComment-695453 Share on other sites More sharing options...
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