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PHP - Insert A Image In Mysql
How can i insert a image in a database and in my images folder using php?
Similar TutorialsOkay, so I found this code on-line that uploads images via a form, inserts them in to MySQL and displays the images on the page. It works great! Problem is, I spent a long time hacking this script to make it work with my site just to realize it did not re-size the image after it uploaded (dumb me). I have found other code that do this but I really do not want to start over at this point and my level of expertise with PHP are a step below what I need to figure this out. I am attaching the "original" code. If anyone knows how to add the necessary lines of code to make every image re-size (reduce) itself. I would very much appreciate it. Thanks <?php $con=mysql_connect("localhost", "root", "rootwdp")or die("cannot connect"); mysql_select_db("student",$con)or die("cannot select DB"); if(isset($_POST['upload'])) { $img=$_FILES["image"]["name"]; foreach($img as $key => $value) { $name=$_FILES["image"]["name"][$key] ; $tname=$_FILES["image"]["tmp_name"][$key]; $size=$_FILES["image"]["size"][$key]; $oext=getExtention($name); $ext=strtolower($oext); $base_name=getBaseName($name); if($ext=="jpg" || $ext=="jpeg" || $ext=="bmp" || $ext=="gif"){ if($size< 1024*1024){ if(file_exists("upload/".$name)){ move_uploaded_file($tname,"upload/".$name); $result = 1; list($width,$height)=getimagesize("upload/".$name); $qry="select id from img where `img_base_name`='$base_name' and `img_ext`='$ext'"; $res=mysql_fetch_array(mysql_query($qry)); $id=$res['id']; $qry="UPDATE img SET `img_base_name`='$base_name' ,`img_ext`='$ext' ,`img_height`='$height' ,`img_width`='$width' ,`size`='$size' ,`img_status`='Y' where id=$id"; mysql_query($qry); echo "Image '$name' updated<br />"; } else{ move_uploaded_file($tname,"upload/".$name); $result = 1; list($width,$height)=getimagesize("upload/".$name); $qry="INSERT INTO `img`(`id` ,`img_base_name` ,`img_ext` ,`img_height` ,`img_width`, `size` ,`img_status`)VALUES (NULL , '$base_name', '$ext', '$height', '$width', '$size', 'Y');"; mysql_query($qry); echo "Image '$name' uploaded<br />"; } } else{ echo "Image size excedded.<br />File size should be less than 1Mb<br />"; } } else{ echo "Invalid file extention '.$oext'<br />"; } } } function getExtention($image_name){ return substr($image_name,strrpos($image_name,'.')+1); } function getBaseName($image_name){ return substr($image_name,0,strrpos($image_name,'.')); } ?> <!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> <script type="text/javascript"> function addItems() { var table1 = document.getElementById('tab1'); var newrow = document.createElement("tr"); var newcol = document.createElement("td"); var input = document.createElement("input"); input.type="file"; input.name="image[]"; newcol.appendChild(input); newrow.appendChild(newcol); table1.appendChild(newrow); } function remItems() { var table1 = document.getElementById('tab1'); var lastRow = table1.rows.length; if(lastRow>=2) table1.deleteRow(lastRow-1); } </script> <style type="text/css"> <a class="tooltip" and them url href="/http.www.anyurl.com" </a> a.tooltip:hover span{display:inline; position:absolute; border:2px solid #cccccc; background:#efefef; color:#333399;} a.tooltip span {display:none; padding:2px 3px; margin-left:8px; width:150px;} </style> </head> <body> <form method="post" action="" enctype="multipart/form-data"> <table align="center" border="0" id="tab1"> <tr> <td width="218" align="center"><input type="file" name="image[]" /></td> <td width="54" align="center"><img src="Button-Add-icon.png" alt="Add" style="cursor:pointer" onclick="addItems()" /></td> <td><img src="Button-Delete-icon.png" alt="Remove" style="cursor:pointer" onclick="remItems()" /></td> </tr> </table> <table align="center" border="0" id="tab2"> <tr><td align="center"><input type="submit" value="Upload" name="upload" /></td></tr> </table> </form> <table border="0" style="border:solid 1px #333; width:800px" align="center"><tr><td align="center"> <iframe style="display:none" name="if1" id="if1"></iframe> <? $qry="select * from img where img_status='Y' order by id"; $res=mysql_query($qry); $i=0; if(mysql_num_rows($res)){ ?> <div align="center"><ul style="width:650px; border: 0px"> <? while($fetch=mysql_fetch_array($res)){ $hratio=120/$fetch['img_height']; $wratio=120/$fetch['img_width']; $ratio=($hratio < $wratio) ? $hratio : $wratio; $hth=$fetch['img_height']*$ratio; $wth=$fetch['img_width']*$ratio; ?> <li style="width:120px; height:180px; border:0px solid #333;float:left;list-style:none outside none; padding-right:5px;"><img src="upload/<? echo $fetch['img_base_name'].'.'.$fetch['img_ext']; ?>" width="<? echo $wth; ?>" height="<? echo $hth; ?>" title="<? echo "image : ".$fetch['img_base_name'].".".$fetch['img_ext']; ?>" /><br /> <? if($i==0) $fp=fopen("fileInfo.txt",'w'); else $fp=fopen("fileInfo.txt",'a'); fwrite($fp,"Image : ".++$i ."\r\n"); fwrite($fp,"Name : ".$fetch['img_base_name'].".".$fetch['img_ext']."\r\n"); fwrite($fp,"width X height : ".$fetch['img_width']." X ".$fetch['img_height']."\r\n"); fwrite($fp,"Size : ".round($fetch['size']/1024,1)."Kb\r\n"); fwrite($fp,"____________________________________\r\n"); fclose($fp); echo $fetch['img_base_name'].".".$fetch['img_ext'].'<br />'; echo $fetch['img_width'].' X '; echo $fetch['img_height'].'<br />'; echo round($fetch['size']/1024,1) .'Kb'; ?> </li> <? }?> </ul> </div><? }?> </td></tr></table> </body> </html> Can anyone tell me why this is not INSERTing? My array data is coming out just fine.. I've tried everything I can think of and cannot get anything to insert.. Ahhhh! <?php $query = "SELECT RegionID, City FROM geo_cities WHERE RegionID='135'"; $results = mysqli_query($cxn, $query); $row_cnt = mysqli_num_rows($results); echo $row_cnt . " Total Records in Query.<br /><br />"; if (mysqli_num_rows($results)) { while ($row = mysqli_fetch_array($results)) { $insert_city_query = "INSERT INTO all_illinois SET state_id=$row[RegionID], city_name=$row[City] WHERE id = null" or mysqli_error(); $insert = mysqli_query($cxn, $insert_city_query); if (!$insert) { echo "INSERT is NOT working!"; exit(); } echo $row['City'] . "<br />"; echo "<pre>"; echo print_r($row); echo "</pre>"; } //while ($rows = mysqli_fetch_array($results)) } //if (mysqli_num_rows($results)) else { echo "No results to get!"; } ?> Here is my all_illinois INSERT table structu CREATE TABLE IF NOT EXISTS `all_illinois` ( `state_id` varchar(255) NOT NULL, `city_name` varchar(255) NOT NULL ) ENGINE=MyISAM DEFAULT CHARSET=latin1; Here is my source table geo_cities structu CREATE TABLE IF NOT EXISTS `1` ( `CityId` varchar(255) NOT NULL, `CountryID` varchar(255) NOT NULL, `RegionID` varchar(255) NOT NULL, `City` varchar(255) NOT NULL, `Latitude` varchar(255) NOT NULL, `Longitude` varchar(255) NOT NULL, `TimeZone` varchar(255) NOT NULL, `DmaId` varchar(255) NOT NULL, `Code` varchar(255) NOT NULL ) ENGINE=MyISAM DEFAULT CHARSET=latin1; Hello, I'm having a bit of a problem here, all help to this issues would be much appreciated I am trying to use text boxes to insert numbers into the database based on what is inputed. If I have a string, like this for example: $variable = 09385493; And I want to insert it into the database like this: mysql_query("INSERT INTO integers(number) VALUES ('$variable')"); When checking the integers table in my database, looking at the number field, the $variable that was inserted is outputted as 9385493 Notice the number zero was taken out of the front of the number. If the number is double 0's (009385493), both of those zero's would disappear, too. Thanks I need help badly! What I want to do is insert into database the value from the selected radio group buttons.. All of them. There are 10 radio groups total (they can be less, but not more). Thanks! Code: [Select] <?php require_once('Connections/strana.php'); mysql_select_db($database_strana, $strana); ?> <link href="css/styles.css" rel="stylesheet" type="text/css" /> <table width="100%" height="100%" style="margin-left:auto;margin-right:auto;" border="0"> <tr> <td align="center"> <form action="" method="post" enctype="multipart/form-data" name="form1"> <table> <?php $tema = mysql_query("SELECT * from prasanja where tip=2")or die(mysql_error()); function odgovor1($string) { $string1 = explode("/", $string); echo $string1[0]; } function odgovor2($string) { $string1 = explode("/", $string); echo $string1[1]; } while ($row=mysql_fetch_array($tema)) { $id=$row['prasanje_id']; $prasanje=$row['prasanje_tekst']; $tekst=$row['odgovor']; ?> <tr> <td> </td> </tr> <tr> <td class="formaP"> <?php echo $prasanje?> </td> </tr> <tr> <td class="formaO"> <p> <label> <input type="radio" name="Group<?php echo $id?>" value="<?php odgovor1($tekst) ?>" /> <?php odgovor1($tekst) ?></label> <br /> <label> <input type="radio" name="Group<?php echo $id?>" value="<?php odgovor2($tekst) ?>" /> <?php odgovor2($tekst) ?></label> <br /> </p></td> </tr> <tr> <td> <br /> </td> </tr> <?php } ?> </table> <input align="left"type="submit" name="submit" value="Внеси" > </form> </td> </tr> </table> prasanje = question tekst/odgovor = answer The answer table: id - primary question_id - the questions ID whose answer is selected in the radio group user_id - cookie takes care of this answer - the value from radio group date - automatic I have this code: <?php $con = mysql_connect("localhost","hhh","hhh"); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("hhh", $con); // -------------------- // Avatar insert check // -------------------- session_start(); $name = $_POST[name]; $group = $_POST[group]; $age = $_POST[age]; $usernameid = $_SESSION[id]; $result = mysql_query("SELECT * FROM avatars WHERE name='$_POST[name]'"); $num = mysql_numrows($result); if ($num == 0) { mysql_query("INSERT INTO avatars (id, usernameid, name, group, age, xp) VALUES ('', '$usernameid', '$name', '$group', '$age', '0')"); header( 'Location: me/' ) ; } else echo 'Sorry, please pick a new name'; ?> And it does everything but put the data into the datebase. If I add a session befor and after '$request' they both run, but the sql doesn't. No error returns, if just redirects to the other page. Any help? well this is truely embarrising...i have a insert statement which works within phpmyadmin but when using mysqli_query it returns a error.
INSERT INTO users (username, timestamp) VALUES ('test', UTC_TIMESTAMP())
Unknown column 'timestamp' in 'field list'
i've been playing about with this for a few hours now ...tried changing the column name (timestamp), adding ` around column names as well as table name.
the column exists which is the strangest part, and ive even checked there is no space after the column name in the db.
whats going on please?
Hi guys I have a registration form working fine, my database is as below: userid username password repeatpassword I have added another column which is "name", users can update their profile once they have logged in so I have created updateprofile.php and when I login-->go to update profile and insert my name nothing adds to mysql name column this is my code below: <?php include ("global.php"); //username session $_SESSION['username']=='$username'; $username=$_SESSION['username']; //welcome messaage echo "Welcome, " .$_SESSION['username']."!<p>"; if ($_POST['register']) { //get form data $name = addslashes(strip_tags($_POST['name'])); $update = mysql_query("INSERT INTO users (name) VALUES ('$_POST[name]') WHERE username='$username'"); } ?> <form action='updateprofile.php' method='POST'> Company Name:<br /> <input type='text' name='name'><p /> <input type='submit' name='register' value='Register'> </form> can you please tell me where in this code is wrong? Im new in php so please excuse me if I have silly mistakes. thanks in advance I don't understand where the empty value is. I've substituted the variables for text and still have the same problem. Code: Code: [Select] $sql = "INSERT INTO courses (course#, name, subject, semester, ap)VALUES('$courseNum', '$courseName', '$subject', '$semester', '$ap')"; 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 '' at line 1 The Script:
<form method="post" action="<?php echo $_SERVER['PHP_SELF'] ?>">
<input type="text" name="hashtags" />
<input type="submit" name="submit" />
</form>
<?php
if(isset($_POST['submit'])){
$hashtags = explode(", ", $_POST['hashtags']);
// Prints e.g.: Array ( [0] => #tag1 [1] => #tag2 [2] => #tag3 [3] => #tag4 )
print_r($hashtags);
}
?>
This gets inserted into the input field:
#tag1, #tag2, #tag3, #tag4I am looking to check if any of the hashtags inside the array already exist in the database, if it does not exists it should create the new ones in the table. I know how to do this if all do not exists in the array and then it goes over to the MySQL query and inserts all of them. My Question Is: How to insert only the ones which do not exists out of the array, so the ones which do exists do not get inserted again into the table? Edited by glassfish, 14 October 2014 - 10:15 AM. My hosting service has magic_quotes_gpc = On. I was working on my home test server and the following script worked perfectly. Turns out I had magic_quotes_gpc = Off .. I set magic_quotes_gpc = On and restarted. Now the script isn't working. See code and output below. I know something isn't being escaped properly, but I have no clue how/what. Even if I copy and paste the $insert output directly to phpmyadmin, it returns the same error. Code: [Select] //HTML Vars $firstName = $_POST['firstName']; $lastName = $_POST['lastName']; $email = $_POST['email']; $desc = $_POST['desc']; //This is a textarea with long description. $year = $_POST['date']; //MySQL - no connection issues $link = mysql_connect('localhost', '__uesr__', '__passwd__*'); $db = mysql_select_db('__DB__', $link); $insert = "INSERT INTO images (firstName, lastName, email, descript, dateYear) VALUES ('$firstName' , '$lastName' , '$email' , '$desc' , '$year' "; $query = mysql_query($insert); if (!$query) { die ('Can\'t query ' . mysql_error()); } echo $insert; ::OUTPUTS:: Can't query 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 INSERT INTO images (firstName, lastName, email, descript, dateYear) VALUES ('this' , 'is' , 'the@email.com' , 'and. the. description won\'t work.' , '3456' Obviously I am a novice. I have tried using mysql_real_escape_string with and without stripslashes, but I'm not getting anywhere except more errors. Any help would be greatly appreciated. And I don't care about SQL injection AT ALL. I just want the thing to work with proper escaping for the description if a user inputs special chars. Hi I have a XML file as follows: <?xml version="1.0"?> <inspection_form> <inspection_type> <inspection_area_tlb>yard_and_lot</inspection_area_tlb> <inspection_area>Yard and Lot</inspection_area> <items> <item> <item_name>PID Signage/unauthorized sign on pole</item_name> <item_value>0</item_value> </item> <item> <item_name>Landscape well maintained</item_name> <item_value>0</item_value> </item> </items> </inspection_type> <inspection_type> <inspection_area_tlb>pump_island</inspection_area_tlb> <inspection_area>Pump Island and Canopies</inspection_area> <items> <item> <item_name>pumps clean and free of dirt</item_name> <item_value>0</item_value> </item> <item> <item_name>Approved trash cans/clean</item_name> <item_value>0</item_value> </item> </items> </inspection_type> </inspection_form> I want to insert into DB as follows: inspection_area_tlb inspection_area item_name item_value yard_and_lot yard and Lot PID Signage/unauthorized sign on pole 0 yard_and_lot yard and Lot Landscape well maintained 0 pump_island Pump Island and Canopies pumps clean and free of dirt 0 pump_island Pump Island and Canopies Approved trash cans/clean 0 I have written some php code. But every item node as insert for every 'inspection_type'. This is my code $filename="sample.xml"; if(filesize($filename)>0) { $oDOM = new DOMDocument(); $oDOM->loadXML(file_get_contents($filename)); foreach ($oDOM->getElementsByTagName('inspection_type') as $oBookNode) { foreach ($oDOM->getElementsByTagName('item') as $itmNode) { $sSQL = sprintf( "INSERT INTO inspections_master_tablename_import (INSPECTION_TYPE_DB_C_NAME, INSPECTION_TYPE_C_NAME, INSPECTION_TYPE_ITEM_C_NAME,INSPECTION_TYPE_ITEM_VALUE_C_NAME) VALUES ('%s', '%s', '%s', '%s')", mysql_real_escape_string($oBookNode->getElementsByTagName('inspection_area_tlb')->item(0)->nodeValue), mysql_real_escape_string($oBookNode->getElementsByTagName('inspection_area')->item(0)->nodeValue), mysql_real_escape_string($itmNode->getElementsByTagName('item_name')->item(0)->nodeValue), mysql_real_escape_string($itmNode->getElementsByTagName('item_value')->item(0)->nodeValue) ); $rResult = mysql_query($sSQL); if(mysql_errno() > 0) { printf( '<h4 style="color: red;">Query Error:</h4> <p>(%s) - %s</p> <p>Query: %s</p> <hr />', mysql_errno(), mysql_error(), $sSQL ); } } } } Can anyone help me pls. Hi guys, I have an array: array ( [apple]=> red, [orange] => orange, [banana] => yellow ) That I want to insert into the apidata table, the PDO connection is good so I guess my code is wrong:
$mykeys = implode(', ', array_keys($newarr));
$myplaceholders[] = '(' . implode (", ", array_fill(0, count($newarr), '?')) . ')';
$values = array_values($newarr);
$res = $db->prepare("INSERT INTO apidata (item, value) VALUES $myplaceholders") ;
$res->execute([$mykeys, $values]);
The apidata table has three fields id (auto_increment) item and value, where am I going wrong? Thanks I've got Code: [Select] for ($i=1; $i<=5; $i++) { if(isset($_POST['partsusedqty'.$i]) && $_POST['partsusedqty'.$i] != "" && $_POST['partsusedqty'.$i] != "0.00") { mysql_query("INSERT INTO partsused (ptnumber, partqty, partdesc, partprice) VALUES ($ticket, '$partsusedqty'.$i, '$partsuseddesc'.$i, '$partsusedprice'.$i)") or die(mysql_error()); } } I need to know the correct formatting to put these variable variables as values in the mysql query. With this particular code, I get the 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 '.1, ''.1, ''.1)' at line 2" I've tried formatting this an endless number of ways, but I used this particular example because its the one I really thought should work. Everything I've tried that doesn't throw an error put the $partsusedqty in the partqty, partdesc, and partprice fields. Thanks for any help! I have a Form on registration.html through which i trying to get data in mysql through the below php script but there is and mysql syntax error please help me with the below code. Code: [Select] <?php $conn = mysql_connect("localhost", "onlinewe_meghraj", "password123") or die(mysql_error()); $db = mysql_select_db("onlinewe_college") or die(mysql_error()); $name1 = $_POST['name1']; $name2 = $_POST['name2']; $year = $_POST['year']; $department = $_POST['deparment']; $group = $_POST['group']; $in_name = $_POST['in_name']; $in_address = $_POST['in_address']; $phone = $_POST['phone']; $email = $_POST['email']; $mobile1 = $_POST['mobile1']; $mobile12 = $_POST['mobile2']; $comment = $_POST['comment']; $result=mysql_query("INSERT INTO register (name1, name2, year, department, group, in_name, in_address, phone, email, mobile1, mobile2, date, comment) VALUES ('$name1', '$name2', '$year', '$department', '$group', '$in_name', '$in_address', '$phone', '$email', '$mobile1', '$mobile2', '".date("Y-m-d h:i:s")."', '$comment')") or die("Insert Error: ".mysql_error()); echo "REGISTRATION DONE"; ?> Please reply. Thank you. I have a mysql insert statement generated with php that is not populating the table. I've echoed the statement and if I copy and paste into phpmyadmin it works fine. The result of the mysql_query function is true. I've emptied the table so there are no primary key conflicts. I've put the statement in a try catch and it does not display a exception. What else can I try? Here's the statement INSERT INTO `wp_term_relationships` (object_id, term_taxonomy_id, term_order) VALUES (1597,83,0) Works absolute fine if I copy and paste into phpmyadmin. Does not populated the table if run through mysql_query guys, im having a problem here thats making me crazy im making a system and i want to insert the current date in the mysql. I have the field called 'data' in the DB, but when i make the code to insert all the other fields, including the 'data', its work perfectly... unless that damn date! $query = "INSERT INTO news (id, titulo, mensagem, data) VALUES (NULL, '$titulo', '$mensagem', 'date(\"d/m/Y\")')"; whats wrong with that? its stores in DB as '0000-00-00'. I am trying to create a script that takes information from a form and puts in a database. In the action page, I decided to post a URL that shows the user there story that they posted. This is where I ran into the problem. . I realized that for the optional fields I could not just use a seperate insert statement, because this creates a new row. So I desided to use update statments, but this STILL does not work, they values or simply not getting inserted. Here is the code: Code: [Select] <?php if(isset($_POST['hidden'])) { die('SPAM BOT!'); } if ( !isset($_POST['title']) && !isset($_POST['summary']) && !isset($_POST['story']) && !isset($_POST['rating']) && !isset($_POST['cat']) ) { die("<div id='impor'>You forgot to enter one(or more) of the following fields <br /> 1. Title <br /> 2. Summary <br /> 3. Story<br /> </div> "); } mysqlConnect(); //take data from form an\ put them in variable $title_form = bb(mysql_real_escape_string($_POST['title'])); //required $summ_form = bb(mysql_real_escape_string($_POST['summary']));// required $story_form = bb(mysql_real_escape_string($_POST['story'])); $cat_form = $_POST['cat']; $rating_form = $_POST['rating']; $username = $_SESSION['user']; // Make the other var into a list of links mysql_query(" INSERT INTO story_info (title, sum, story, user, cat, rating) VALUES('$title_form','$summ_form', '$story_form,', '$username', '$cat_form','$rating_form') "); if(isset($_POST['notes'])) { $notes_form = mysql_real_escape_string($_POST['notes']); $notes_final = bb($notes_form); mysql_query(" UPDATE story_info SET notes = '$notes_final' WHERE story = '$story_form' AND user = '$username' AND sum = '$summ_form' AND title = '$title_form' "); } //put other in array. Use while loop to put link code. Then but it back into one non array variable if(isset($_POST['u_id'])) { $uid = mysql_real_escape_string($_POST['u_id']); $uid_db = str_replace(' ','_', $uid); $blerg = " UPDATE story_info SET series_id = '$uid_db' WHERE story = '$story_form' AND user = '$username' AND sum = '$summ_form' AND title = '$title_form' "; mysql_query($blerg); } echo "<h1> Your Story Has Been Posted! Thanks for posting $username . </h1>"; echo "Please review the post below <br />"; echo "<h2> $title_form </h2>"; echo "<strong> <h2> Summary: </h2> </strong> $summ_form"; echo "<h4> Story: </h4>"; echo "$story_form"; if(isset($notes)) { echo "<h4> Author's Notes: </h4> "; echo "$notes_final"; } if (isset($uid_db)) { echo '<h3> Unique Series ID </h3>'; echo '<p> Make sure to write down this! <br />' .$uid_db .'</p> '; } $db = mysql_query(" SELECT story_id FROM story_info WHERE story='$story_form' AND user='$username' ")or die(mysql_error()); $rows = mysql_fetch_assoc($db); $id = $rows['story_id']; echo "Catagory: $cat_form <br /> Rating: $rating_form <br /> "; echo "<a href='?p=page&id=$id'> Click here to view your story! </a>'"; ?> Please help!
Array ( [data] => Array ( [0] => Array ( [latitude] => 22.934566 [longitude] => 79.08728 [type] => county [distance] => 44.328 [name] => Narsinghpur [number] => [postal_code] => [street] => [confidence] => 0.5 [region] => Madhya Pradesh [region_code] => MP [county] => Narsinghpur [locality] => [administrative_area] => [neighbourhood] => [country] => India [country_code] => IND [continent] => Asia [label] => Narsinghpur, India ) ) ) This is probably some obvious error I have made, but I cannot figure it out. I have made a few pages and now I am debugging them. My first page is called insert_purchase_order.php; on this page a person will enter some data in fields and hit the insert button. Then, the data is passed to another page, but when I try to insert into mysql it does not give me any errors, but I have no new rows either. The code for my 2nd page: Code: [Select] <?php session_start(); $action=$_GET[action]; if ($action==insert){ $randid=$_POST['randid']; $vendor=$_POST["vendor"]; $purchase_order_date=$_POST["purchase_order_date"]; $ship=$_POST["ship"]; $fob=$_POST["fob"]; $terms=$_POST["terms"]; $buyer=$_POST["buyer"]; $freight=$_POST["freight"]; $req_date=$_POST["req_date"]; $confirming_to=$_POST["confirming_to"]; $remarks=$_POST["remarks"]; $tax=$_POST["tax"]; $con = mysql_connect("localhost","root","pass"); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("main", $con); mysql_query("INSERT INTO purchase_order (randid, vendor, purchase_order, ship, fob, terms, buyer, freight, req_date, confirming_to, remarks, tax) VALUES ($randid, $vendor,$purchase_order_date,$ship, $fob, $terms, $buyer, $freight, $req_date, $confirming_to, $remarks, $tax)"); mysql_close($con); echo 'Data Accepted...'; echo '<br/>'; echo 'P.O. Inserted Successfully'; }else{ echo 'Error... Please Contact Bruce.'; echo 'Bruce, no data was passed from the insert_purchase_order.php page.'; } ?> <a href="http://localhost/insert_purchase_order_items.php?po= <?php echo $randid; ?>">Insert Purchase Order Items</a> I have permissions and everything. Thanks |
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