PHP - Php With Mysql Or Javascript With A Text File
This is my website http://www.rahulkadukar.info/
As you can see I have the blue navigation bar on top and on mouse hover you can see a drop down menu. The contents of the drop down menu can change and I wanted to write a script that would populate the contents of the drop down menus. Which of these is a better option 1. Use JavaScript along with a text file and use JavaScript to populate the drop down 2. Use a SELECT in PHP to populate the drop down menu I am expecting that the 10 main columns would have at the most 20 entries each. Similar TutorialsI am trying to add a txt file to a mysql database. Is there any one that can help? Here are the details TEST.TXT contains the following A001200910019999999900000200000000000000000000000 00000000000000000000000000 A002200910019999999900000610000000000000000000000 00000000000000000000000000 A003200910019999999900000687500000000000000000000 00000000000000000000000000 A004200910019999999900000335000000000000000000000 00000000000000000000000000 A005200910019999999900000626500000000000000000000 00000000000000000000000000 A006200704019999999900000423500000000000000000000 00000000000000000000000000 A007200910019999999900000323500000000000000000000 00000000000000000000000000 A008200704019999999900000102500000000000000000000 00000000000000000000000000 And so on up to 200 rows long I need to split each row the same way and add them to a table I will use the first row as an example Character 1 to 4 (A001) entered into a field called name Character 5 to 12 (20091001) entered into a field called efdate Character 13 to 20 (99999999) entered into a field called exdate Character 21 to 31 (00000200000) entered into a field called prfee Character 32 to 42 (00000000000) entered into a field called assfee Character 43 to 53(00000000000) entered into a field called spfee Character 54 to 64 (00000000000) entered into a field called anfee Character 65 to 75 (00000000000) entered into a field called nanfee Thanks in advance for the help Hello all brilliant minds, I'm a new in all this world of DB and coding and always i tried to avoid it because I think it is very hard (I'm trying to change now). problem: I have a text file (log.txt) have data like below ======================================== > rtrv-ls Command Accepted - Processing OXX 12-02-21 08:44:41 EST EAGLE5 42.0.1-63.38.31 rtrv-ls Command entered at terminal #7. ; OXX 12-02-21 08:44:41 EST EAGLE5 42.0.1-63.38.31 L3T SLT GWS GWS GWS LSN APCI (SS7) SCRN SET SET BEI LST LNKS ACT MES DIS SLSCI NIS gtcen1pls 7-010-4 none 1 2 no D 2 off off off no off gtgdv1pls 7-010-5 none 1 2 no D 2 off off off no off > rtrv-sid Command Accepted - Processing OXX 12-02-21 08:43:43 EST EAGLE5 42.0.1-63.38.31 rtrv-sid Command entered at terminal #7. ; OXX 12-02-21 08:43:43 EST EAGLE5 42.0.1-63.38.31 PCA PCI PCN CLLI PCTYPE 010-010-010 7-055-1 01830 crher1p OTHER ; . . may output like this starting always with > =================================================== Then I have SQL DB that have column with "rtrv-ls" and "rtrv-sid" Requirement : A) Use Php to open the .txt file, I used HTML form so the user can upload the file on the browser then B) Search for the column name as pattern C)Send query to MYSQL to update DB with Data for each column D) Query the Mysql for several possible outputs (I've done this part using HTML form and simple php code as below. <?php // If we got a GET request on the page from HTML part of the code, we will want to store the data in a variable e.g $my_country // So country will be passed via GET and will be stored in $my_country variable. $host="localhost"; $con_usr="aomar"; $con_pass="nokia"; if ($_POST) { $user_input = $_POST['my_log'] ; // We use the request to search in Database and return the result, First step we need to connect to the database // $connect = mysql_connect ($host,$con_usr,$con_pass); // Then we need to select database to run the query and return the result the connection stored in the $connect ressource // if ($connect) { mysql_select_db("eagle",$connect); // We store the query in variable $query becuase is it quite long text; we actually attach our varaiable $my_log passed from HTML // $query = "SELECT `".$user_input."` FROM `ss7`"; //$update="INSERT INTO ss7 ('".$user_input."') VLAUES ('; // SO if user selected Egypt it will be stored in (name=my_country) variable from HTML and will be send via GET // Then we store the query results in a variable called $result $result = mysql_query ($query) ; $arr= mysql_fetch_array($result); echo nl2br ($arr["$user_input"]); //Finally we return what we want from the array. // Print the output as we wish. in case of row data like STP output, use the nl2br to preserve the output as it was written to MYSQL DB // We need to check if the Query return no value (!$arr) , and print corresponding action } //================SECONED PART WHEN I LOAD THE LOG FILE TO BE STORED IN MYSQL================// //close the db mysql_close ($connect); } ?> <html> <body> <form action = "eagle.php?pmode=my_file" method = "POST" enctype="multipart/form-data> <fieldset> <legend> Enter you log file here and click Save: </br> </legend> <label for="my_file"> This Is My Capture File </label> <input type = "file" name ="my_file" style=margin:auto > </input><br/><br/> <input type="submit" value="Upload file"> </fieldset> </form> </body> </html> <html> <body> <title>Eagle Data Base Store</title> <form action = "eagle.php?pmode=my_log" method = "POST"> <fieldset> <legend> Please Select the DB you would like to retrive </legend> <br/><br/> <input type = "radio" name ="my_log" value = "rtrv-serial-num">Serial-Num</input> <input type = "radio" name ="my_log" value = "rtrv-stp">rtrv-stp</input> <input type = "radio" name ="my_log" value = "rtrv-feat">rtrv-feat</input> <input type = "radio" name ="my_log" value = "rtrv-dstn">rtrv-dstn</input> <input type = "radio" name ="my_log" value = "rtrv-rte">rtrv-rte</input> <input type = "radio" name ="my_log" value = "rtrv-trm">rtrv-trm</input> <input type = "radio" name ="my_log" value = "rtrv-slk">rtrv-slk</input><br/> <br/> <input type = "radio" name ="my_log" value = "rtrv-ls">rtrv-ls</input> <input type = "radio" name ="my_log" value = "rtrv-sccpopts">rtrv-sccpopts</input> <input type = "radio" name ="my_log" value = "rtrv-stpopts">rtrv-stpopts</input> <input type = "radio" name ="my_log" value = "rtrv-gsmopts">rtrv-gsmopts</input> <input type = "radio" name ="my_log" value = "rtrv-cmd">rtrv-cmd</input> <input type = "radio" name ="my_log" value = "rtrv-assoc">rtrv-assoc</input><br/> <br/> <input type = "radio" name ="my_log" value = "rtrv-gpl">rtrv-gpl</input> <input type = "radio" name ="my_log" value = "rtrv-ip-host">rtrv-ip-host</input> <input type = "radio" name ="my_log" value = "rtrv-ip-lnk">rtrv-ip-lnk</input> <input type = "radio" name ="my_log" value = "rtrv-secu-trm">rtrv-secu-trm</input> <input type = "radio" name ="my_log" value = "rtrv-secu-user">rtrc-secu-user</input> <input type = "radio" name ="my_log" value = "rept-stat-sys">rept-stat-sys</input><br/><br/> <input type = "radio" name ="my_log" value = "rept-stat-clk">rept-stat-clk</input> <input type = "radio" name ="my_log" value = "rtrv-ctrl-feat">rtrv-ctrl-feat</input> <input type = "radio" name ="my_log" value = "rtrv-tabl-capacity">rtrv-tabl-capacity</input><br/><br/> <input type = "submit" value = "Show Selected Table"/> </fieldset> </form> </body> </html> ==========================================END OF CODE=========================== *** Sorry for the many comments but I want to be sure I remember why I do that . You help is much appreciated (Please note that need to understand more than I need the Code itself) This topic has been moved to JavaScript Help. http://www.phpfreaks.com/forums/index.php?topic=320945.0 create table mimi (mimiId int(11) not null, mimiBody varchar(255) ); <?php //connecting to database include_once ('conn.php'); $sql ="SELECT mimiId, mimiBody FROM mimi"; $result = mysqli_query($conn, $sql ); $mimi = mysqli_fetch_assoc($result); $mimiId ='<span>No: '.$mimi['mimiId'].'</span>'; $mimiBody ='<p class="leading text-justify">'.$mimi['mimiBody'].'</p>'; ?> //what is next? i want to download pdf or text document after clicking button or link how to do that Hi, I am writing several scripts and some are used to amend extra information to a text file. However, I added a hyperlink to the text file so that the user can go back to a page where they can add extra information. However, since I have done this every time I amend more text to the text file, the extra text appears below the hyperlink rather than above it, and I was wondering if there was a way around this. My amend code is as follows: Code: [Select] <html> <head> <title>Amend File</title> <link rel="stylesheet" type="text/css" a href="rcm/stylesheet.css"> </head> <?php if($_POST['append'] !=null) { $filename="C:/xampp/htdocs/rcm/denman2.txt"; $file=fopen($filename, "a"); $msg="<p>Updated Information: " .$_POST['append']. "</p><br>"; fputs ($file, $msg); fclose($file); } ?> <body> <h1>Do you want to append to a document?</h1> Enter Updated Information: <form action="amendfile2.php" method="post"> <input type="text" size="40" name="append"><br><br> <input type="submit" value="Add updated information to report"> </form> <form action="viewfile3.php" method="post"> <input type="submit" size="40" value="View Web Blog"> </form> <form action="loginform.php" method="post"> <input type="submit" value="Click here to go to the Log In Screen"> </form> </body></html> And my text file is as follows: Code: [Select] <h1>Accident Report</h1> <p>First Name: Andrew Last Name: Denman Age: 18 Complete Weeks Since Accident: 2<br> <a href="amendfile2.php">Amend to this file</a> Any help would be appreciated Ok i have been working on this for a day+ now. here is my delema simple .ini text file. when a user makes a change (via html form) it makes the correct adjustments. problem is the newline issue 1. if i put a "\n" at the end (when using fputs) works great, except everytime they edit the file it keeps adding a new line (i.e. 10 edits there are now 10 blank lines!!!!) 2. if i leave off the "\n" it appends the next "fgets" to that lilne making a mess Code: [Select] ##-- Loop thruoght the ORIGINAL file while( ! feof($old)) { ##-- Get a line of text $aline = fgets($old); ##-- We only need to check for "=" if(strpos($aline,"=") > 0 ) { ##-- Write NEW data to tmp file fputs($tmp,$info[$i]." = ".$rslt[$i]."\n"); $i++; } ##-- No Match else { fputs($tmp,$aline."\n"); }//Checking for match }//while eof(old) what in the world is making this such a big deal. i dont remember having this issue in the past I tried opening with w+, and just w on the temp file a typical text line would be some fieldname = some value the scipt cycles through the file ignoring comments that are "#" ps the tmp file will overwrite the origianl once complete all i really want to know is WHY i cant get the newline to work, and what is the suggested fix EDIT: i just tried PHP_EOL and it still appends another newline I currently am working on a project where I code a "simple" telephone directory. There are three main tasks that it needs to do: 1. Directory.php(index page) has a "First Name" and "Last Name" field and a search button. When a name is searched from the directory.txt file, it displays First Name, Last Name, Address, City, State, Zip and phone in findinfo.php in designated text boxes...first name, last name, etc. 2. From the findinfo.php, like previously stated, the users information is listed in the appropriate text boxes. From there, there is an update button that will overwrite the user's information to directory.txt if that button is selected. It will then say the write was sucessful. 3. (completed this step) From the index page, there is a link that will take you to addnew.php where you enter First Name, Last Name, Address, City, State, Zip and phone in a web form and write it to directory.txt. This is the php code for the third step: <?php $newentryfile = fopen("directory.txt", "a+"); $firstname = $_POST['fname']; $lastname = $_POST['lname']; $address = $_POST['address']; $city = $_POST['city']; $state = $_POST['state']; $zip = $_POST['zip']; $phone = $_POST['phone']; $newentry = "$firstname $lastname\n\r $address\n\r $city, $state $zip\n\r $phone\n\r"; if (flock($newentryfile, LOCK_EX)) { if (fwrite($newentryfile, $newentry) > 0) echo "<p>" . stripslashes($firstname) . " " . stripslashes($lastname) . " has been added to the directory.</p>"; else echo "<p>Registration error!</p>"; flock($newentryfile, LOCK_UN); } else echo "<p>Cannot write to the file. Please try again later</p>"; fclose($newentryfile); if(empty($firstname) || empty($lastname) || empty($address) || empty($city) || empty ($state) || empty($zip) || empty($phone)) { echo "<p>Please go back and fill out all fields.</p>"; } ?> So to sum it all up, what would be my best approach? I am totally stumped and not sure which function to use. Should I work my way from step 1 to step 2? I see it as when I do the search for the name from directory.php, it takes me to findinfo.php, listing the users information in the text boxes. From there, if I needed to, having the user's information already listed I could hit the update button to overwrite the new information to directory.txt. Doing the update when then tell me that the write was successful. I have literally been scouring the internet for hours. What would be the best function to do this? I hope I was clear enough. Please help me out and thank you for your time. Hello, i am currently getting an Microsoft Excel formatted text file whose save type is .Txt from a URL.I used to open it and will change the save type as excel file. Please suggest whether we can do this with php code. currently my code is like this, <? php copy("http://www.faa.gov/airports/airport_safety/airportdata_5010/menu/emergencyplanexport.cfm?Region=&District=&State=&County=&City=LAS%20VEGAS&Use=&Certification=","./contactsexport.xls"); ?> where as the contactsexport.xls type is .Txt which i need it in .xls Thanks in Advance. Hey guys i have a problem, I have a form to upload 2 image files however sometimes i want to only upload the one file, but if i only upload one image, it tells me that it can't upload because there is no name for the second image. I tried fixing it with if (isset($fieldname2)) however i just realised that $feildname2 is always set so im not quite sure how to fix? Code: [Select] ($action == "add"){ $add = $_POST['add']; $reference = $_POST['reference']; $category = $_POST['category']; $sub_cat = $_POST['sub_cat']; $fieldname = 'file'; $fieldname2 = 'file2'; error_reporting(E_ALL); ini_set('display_errors',1); is_uploaded_file($_FILES[$fieldname]['tmp_name']); getimagesize($_FILES[$fieldname]['tmp_name']); if (isset($fieldname2)) { is_uploaded_file($_FILES[$fieldname2]['tmp_name']); getimagesize($_FILES[$fieldname2]['tmp_name']); } $now = time(); while(file_exists($uploadFilename = $uploadsDirectory.$_FILES[$fieldname]['name'])) { $now++; } if (isset($fieldname2)) { while(file_exists($uploadFilename2 = $uploadsDirectory.$_FILES[$fieldname2]['name'])) { $now++; } } move_uploaded_file($_FILES[$fieldname]['tmp_name'], $uploadFilename); if (isset($fieldname2)) { move_uploaded_file($_FILES[$fieldname2]['tmp_name'], $uploadFilename2); } $sqlname = $_FILES[$fieldname]['name']; if (isset($fieldname2)) { $sqlname2 = $_FILES[$fieldname2]['name']; } $id = mysql_insert_id(); $query = "INSERT INTO cakes (id, reference, category, sub_cat, image, image2) VALUES ('$id', '$reference', '$category', '$sub_cat','$sqlname','$sqlname2')"; $sql = mysql_query($query) or die (mysql_error()); } echo("<form name='add' method='post' enctype='multipart/form-data' action='?action=add'>"); echo("<input type='hidden' name='MAX_FILE_SIZE' value='3000000"); echo("<input type='hidden' name='?action=add'>"); echo("<table class=main cellspacing=0 cellpadding=2 width=60%>"); echo("<tr><td>Reference: </td><td align='right'><input type='text' size=50 name='reference' value='$reference'></td></tr>"); echo("<tr><td>Category: </td><td align='center'><select name='category'><option>Occasional</option><option>Special</option><option>Figures</option><option>Novelty</option><option>Wedding</option></SELECT></td></tr>"); echo("<tr><td>Sub Category: </td><td align='center'><select name='sub_cat'><option></option><option>Birthday</option><option>Anniversary</option><option>Christmas</option><option>Christening</option><option>Valentine</option><option>Mothers Day</option><option>Fathers Day</option><option>Easter</option></select></td></tr>"); echo("<tr><td>Cake Image: </td><td align='center'><input id='file' type='file' name='file'></td></tr>"); echo("<tr><td>Cake Image 2: </td><td align='center'><input id='file2' type='file' name='file2'></td></tr>"); echo("Before uploading an image please resize the image to 290px x 218px"); echo("<tr><td></td><td><div align='right'><input type='Submit'></div></td></tr>"); echo("</table>"); hello, I need xml file as playlist.xml and i want to retrive data from my mysql table. I have the following code as i am trying to use php code inside playlist.xml file. Code: [Select] <?xml version="1.0" encoding="utf-8"?> <xml> <?php $dbhost='localhost'; $dbuser='pavel'; $dbpass='pavel123'; mysql_connect($dbhost,$dbuser,$dbpass); mysql_select_db("amarmusic"); $sql=("select *from bband where artist='Meghdol' and album='Drohe Montre Valobasha';"); $res = mysql_query($sql); while ($result = mysql_fetch_array($res)){ ?> <track> <path> <? $result['link']?></path> <title><? $result['title']?></title> </track> <? } ?> </xml> but unfortunately its not working. any1 have any idea how to fix it? please help me. Hiya guys After getting everything else working how I expected, I'm sort of struggling on the last step. Downloading I have an upload.php file that allows me to upload a file to Mysql, the fields available a upid - Primary id - Need to link this to the logged on user id name type size content The upload works perfectly Can anyone help with implmenting it to the profile.php (1st page after login) On profile page I have: Welcome "username" from Session Dynamic Table display his user id, username and password at the moment, this will be changed as not needed tho. I am using the sessions MM_Username to pass from the login The table for Login looks like: id - primary username password I assume that if I can copy the ID from login and put it in ID in Upload and add colums to dynamic table to show the upload file, will this make that file only available to logged in user? Cheers I have the code below , and when I run it on my website , it returns a XML Parsing Error: not well-formed error , which says it has a problem reading the & i think
please see http://www.jamesflow...om/php/xml2.php
I think it has something to wiht the str_replace as when I change the $row['customerName'] . "</Name>\n"; to $row['customerNumber'] . "</Name>\n"; as a test it works fine , I understand XML cant handle special characters , how would prevent this happening?
code as below
$query = "SELECT * FROM customers"; $resultID = mysql_query($query, $linkID) or die("Data not found."); $xml_output = "<?xml version=\"1.0\"?>\n"; $xml_output .= "<entries>\n"; for($x = 0 ; $x < mysql_num_rows($resultID) ; $x++){ $row = mysql_fetch_assoc($resultID); $xml_output .= "\t<entry>\n"; $xml_output .= "\t\t<Name>" . $row['customerName'] . "</Name>\n"; // Escaping illegal characters $row['text'] = str_replace("&", "&", $row['text']); $row['text'] = str_replace("<", "<", $row['text']); $row['text'] = str_replace(">", ">", $row['text']); $row['text'] = str_replace("\"", """, $row['text']); $xml_output.= "\t\t<Number>" . $row['customerNumber'] . "</Number>\n"; $xml_output.= "\t</entry>\n"; } $xml_output .= "</entries>"; echo $xml_output; regards James Hi, I am a newbie to php and am trying to copy mysql data to a text file. The format I need to create is Hi,
If I place the following into a file called db.php and save it into folder called includes using xampp:
$dbhost = 'localhost'; Hello, I am storing files that my users upload to the website as a blob in mysql database. Here is the code that does uploading: $fileName = $_FILES['userfile']['name']; $tmpName = $_FILES['userfile']['tmp_name']; $fileSize = $_FILES['userfile']['size']; $fileType = $_FILES['userfile']['type']; $fp = fopen($tmpName, 'r'); $content = fread($fp, filesize($tmpName)); $content = mysql_real_escape_string($content); fclose($fp); if(!get_magic_quotes_gpc()) { $fileName = addslashes($fileName); } $query = mysql_query("INSERT INTO documents (idapplicant, fileType, fileSize, fileData, fileName) VALUES ('$idapplicant','$fileType','$fileSize', '$content','$fileName')",$db); Here is the code for download: echo "File found:".@mysql_result($result, 0, "fileID"); //always correct $data = @mysql_result($result, 0, "fileData"); $name = @mysql_result($result, 0, "fileName"); $size = @mysql_result($result, 0, "fileSize"); $type = @mysql_result($result, 0, "fileType"); header("Content-length: $size"); header("Content-type: $type"); header("Content-Disposition: attachment; filename=$name"); echo $data; I can upload/download files no problem. However, I noticed problem with GIF files. All GIF files are just dark screen after downloading. All other files (PDF, JPEG) looked just fine. I compared files (before upload and after download) in HEX editor and found that the first byte on all files is set to A0 (it is added in front of all the data) and the last byte is deleted! I have tried: $content = addslashes($content); instead of: $content = mysql_real_escape_string($content); with the same results I also tried removing this line of code and in that case no file would be uploaded at all! Any idea? Thank you I need to write a bit of code to select some mp3's (actually, just the titles) from a database...but I need to select them based on last play (which is a timestamp field in the DB) AND I need to select 14 minutes worth of them (and not more than 15 minutes). I currently do NOT know how long each song is... So, if every song was 2 minutes long, I would need 7 songs. If each song was 3 minutes long, I'd need to select 5 of them. I currently do NOT have a length field in the db (the boss sprung this on me after I designed the DB). A few questions: Would it be best to have a column of TIME (mysql time...not the name of the column)? Would it be best to have a column for the raw size of the file (each file is the same bitrate, so I could do the math - plus it would be probably be easier to import raw filesize using php into the db) Would it be easier to do something else? Once we answer that, I'll start with some code...and probably be back for help! Hi, i have a table which shows some mysql data, every entry has a checkbox to select individual entries, now i want to be able to export those selected entries into a xml or txt file, i tried this: Code: [Select] <?php if($_POST['exporttxt']){ for($i=0;$i<count($_POST['checkbox']);$i++){ $export_id = $checkbox[$i]; $sql = "SELECT * FROM table WHERE id='$export_id'"; $result = mysql_query($sql);} $output = "<?xml version=\"1.0\" encoding=\"UTF-8\"?>\n<root>\n"; if($result->num_rows > 0) { while($myrow = $result->fetch_assoc()) { $output .= "\t<row>\n"; foreach($myrow as $_name => $_value) { $output .= "\t\t<$_name>$_value</$_name>\n"; } $output .= "\t</row>\n"; } } $output .= "</root>"; } header('content-type: text/xml'); header('content-disposition: attachment; filename=data_export.xml'); echo $output; exit; ?> but that doesn't seem to work. Any hints ? Hey everyone, I posted something similar to this before and thought I had received an answer, but for whatever reason, the script I wrote had inconsistent functionality. It would work sometimes, but not others. At any rate I had to completely revamp the script. Now I'm having the same initial problem with this new script. Basically, what I'm trying to do is get my PHP script to upload a file, insert form info into the MySQL Database Table, and insert the name of the uploaded file into the MySQL Database table as well. I've been able to get everything to work except inserting the name of the uploaded file into the MySQL Database table. I'll show you what I have currently (a stripped down version anyway) then any suggestions you could give would be much appreciated! Please help! I've been working on this for the past month and keep running into issues: Code: [Select] <?php $target_path = "Images/"; $target_path = $target_path . basename($name = $_FILES['pic']['name']); $first_name=$_POST['first_name']; $last_name=$_POST['last_name']; $middle_init=$_POST['middle_init']; mysql_connect("localhost", "My_Username", "My_Password") or die(mysql_error()) ; mysql_select_db("My_Database") or die(mysql_error()) ; if (!$_POST['first_name'] | !$_POST['last_name'] | !$_POST['middle_init']) { header('Location:http://www.mysite.com/error.html'); die(); } $insert = "INSERT INTO image (first_name, last_name, middle_init, name) VALUES ('".$_POST['first_name']."', '".$_POST['last_name']."', '".$_POST['middle_init']."', '".$_POST['name']."')"; $add_member = mysql_query($insert); move_uploaded_file($_FILES['pic']['tmp_name'], $target_path) ?> <HTML> <HEAD> <TITLE>My Site</TITLE> HTML "Confirmation Script" continues on from there. Like I said before, everything works except the name of the uploaded file doesn't get inserted into the MySQL table. Oh...that specific field in the MySQL table is titled "name". Anyway, again, any help you could give would be awesome! Thanks! Hey guys, first post and I've just returned to PHP after about 4 years of no coding, so be gentle! In a nutshell I'm creating a photo album and I've pretty much got the majority of it complete, apart from a few tweaks and the obvious ongoing development. I'm at the stage now where I need to moderate images being uploaded, so I've made an admin only script which displays the uploaded images with links that say approve and delete. Uploaded images are stored in uploads/ which are left there until i move them to img/, plus the filename is stored in mysql so I can "<img src='../uploads/".$row['filename']."' width='200'>". Now, I would like to make the Approve button move the image from uploads/ to img/ and I'd like the delete button to remove both the entry from MySQL and the file from the uploads folder and I'm not too sure on how to make it work. Here's what I have so far in the mod.php file (mod for moderation) Code: [Select] $image = mysql_query("SELECT * FROM images WHERE id"); while($row = mysql_fetch_assoc($image)) { echo " <table> <td> <tr> <img src='../uploads/".$row['filename']."' width='200'><br /> </tr> </td> <td> <tr> <a href=''>Approve</a> <a href=''>Delete</a> </tr> </td> </table> "; } You'll have to ignore the table tags, I'm still getting used to positioning items on the screen lol. Any clues would be greatly appreciated Live long and prosper. <!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>Import a CSV File with PHP & MySQL</title> </head> <body> <form action="" method="post" enctype="multipart/form-data" name="csv" id="csv"> Choose your file: <br /> <input name="csv" type="file" id="csv" /> <input type="submit" name="Submit" value="Submit" /> </form> </body> |