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PHP - Copy An Array?
hi i made my very first array, and its working great
now i want to have another array wich does the same thing, but on another category, it shows gold silver and bronze logos, but now i want to give leaders of other categorys also those logos, but different ones so i thought i change from $array[$snipergameid][$x] = $sniperleaders[playerid]; (part of the code in a while loop) into $array2[$snipergameid][$x] = $sniperleaders[playerid]; (part of the code in a while loop) and i do the same for the result checker loops but that doesnt work so how do i copy an entire arrray, and give it another name to match it with the correct while loop checker? Similar TutorialsHow would I copy an array of selected files? I know how to list them using a 'for' loop, and how to separate them using a 'foreach' loop, but I am unable to get a checkbox array to copy to a destination folder. Help is appreciated. Hi, I'm not sure if this is a wierd one, I have lots of arrays i'm looping through like this to see them:- for($i=0; $i <count($confirmed); $i++){//loop through print_r_html($confirmed[$i]); } Which shows the following :- Code: [Select] Array ( [0] => AAAAAA-BBBBBB [1] => 2 [2] => 2010-09-29 ) Array ( [0] => AAAAAA-BBBBBB [1] => 2 [2] => 2010-09-28 ) Array ( [0] => CCCCCC-DDDDDD-EEEEEE [1] => 3 [2] => 2010-09-29 ) Key 0 is a Groupconcat split by "-" Key 1 shows how many fields are in 0 Key 2 is just the date What I would like to do is split the first array into 2 arrays, the second in 2, and the third into 3 (therefore ungrouping the groupconcat if you will) as such:- Code: [Select] First into these two:- Array ( [0] => AAAAAA [1] => 2 [2] => 2010-09-29 ) Array ( [0] => BBBBBB [1] => 2 [2] => 2010-09-29 ) Second into these two:- Array ( [0] => AAAAAA [1] => 2 [2] => 2010-09-28 ) Array ( [0] => BBBBBB [1] => 2 [2] => 2010-09-28 ) Third into these two:- Array ( [0] => CCCCCC [1] => 3 [2] => 2010-09-29 ) Array ( [0] => DDDDDD [1] => 3 [2] => 2010-09-29 ) Array ( [0] => EEEEEE [1] => 3 [2] => 2010-09-29 ) ) I know I can get out the info from the first of the keys in each array by putting this into the loop:- for($i=0; $i <count($confirmed); $i++){//loop through if($i==0){ $seperated[] = explode("-",$confirmed[$i]);//split by "-" } } print_r_html($seperated); but i'm not sure how to copy the other keys at the same time? If anyone has got any ideas that would be great! thanks in advance Kev. When will modern PHP copy an array by reference only instead of reassigning to memory? For instance, when I execute $bigArray=$stmt->execute([$id]), PHP utilizes memory to store the data. But then when I execute $dataHolders[] = new DataHolder($bigArray), I believe $bigArray is passed by reference and a new copy of $bigArray will not be stored in memory. If in DataHolder::__construct, I executed $data[]='bla', then it is my understanding that PHP would see that the array is being modified and will automatically make a copy, but I am not doing so and just state this to confirm my understanding. The part that I am uncertain about is not the first time data is retrieved but any additional times. When I execute the next $bigArray=$stmt->execute([$id]), will PHP need to first store in memory the array that DataHolder::data references, then release the memory associated with $bigArray, and then store in memory the new $bigArray?
<?php
class DataHolder
{
private $data;
public function __construct(array $data)
{
$this->data=$data;
//$this->data[]='bla'; <-- I believe this will force PHP to make a new copy of the array but I am not doing so
}
}
$dataHolders=[];
foreach($ids as $id) {
$bigArray=$stmt->execute([$id]);
$dataHolders[] = new DataHolder($bigArray);
}
Ok I am designing a php upload that will take a image file from a form and change the name of the file to the productnumber also recieved from the form. I had it working the otherday now it says Warning: copy() [function.copy]: open_basedir restriction in effect File() is not within the allowed path(s): (/home:/tmp:/usr) addpro.php on line 51. my files are attached.... note that $pnum is the product number gotten from my form and image is the image being uploaded gotten from the form also. The thing is it worked the other day but now it don't is it a change to the server ( I dont run the server) or did I mess up my code since then? I really need a code that will do this two time over once for a small image being put into a folder called small and once for a folder called large both images being uploaded and being changed to $pnum.ext so they will both be displayed when being called out by the product number. but I can work on that after I get this one working. line 51 is $copied = copy($_FILES['image']['tmp_name'], $newname); Hi guys, I am wirting a script to upload an image file, but I get the error: copy() [function.copy]: Filename cannot be empty in Obviouly it has something to do with the copy() function - (copy($HTTP_POST_FILES['product_img']['tmp_name'], $path)) I think it is not copying from my temp folder or something. Can anyone help me out? Code is: Code: [Select] //IMAGE FILE $file_name = $_FILES['product_img']['name']; echo '<p>File Name:' . $file_name . '</p>'; //IMAGE UPLOAD // random 4 digit to add to file name $random_digit=rand(0000,9999); //combine random digit to file name to create new file name //use dot (.) to combile these two variables $new_image_name = $random_digit.$file_name; echo $new_image_name; //SET DESINATION FOLDER $path= "uploads/".$new_image_name; echo '<p>Path:' . $path . '</p>'; if($product_img !=none) { if(copy($HTTP_POST_FILES['product_img']['tmp_name'], $path)) { echo "Successful<BR/>"; echo "File Name :".$new_file_name."<BR/>"; echo "File Size :".$HTTP_POST_FILES['product_img']['size']."<BR/>"; echo "File Type :".$HTTP_POST_FILES['product_img']['type']."<BR/>"; } else { echo "Image upload Error<BR/>"; } } Any help would be greatly appreciated! Thanks This was generated through phpFormGenerator, but I'm getting the following warning message when a file is not uploaded. I didn't set it as a required field. "Warning: copy() [function.copy]: Filename cannot be empty ..." If it possible to skip this check since it's not supposed to be required to upload a file? I'm simply trying to make a contact form that emails basic information with the added option of attaching a file/picture. Code: [Select] <?php include("global.inc.php"); $errors=0; $error="The following errors occured while processing your form input.<ul>"; pt_register('POST','FullName'); pt_register('POST','Email'); pt_register('POST','Phone'); pt_register('POST','QuestionsorComments'); $QuestionsorComments=preg_replace("/(\015\012)|(\015)|(\012)/"," <br />", $QuestionsorComments);$AttachPictureforQuote=$HTTP_POST_FILES['AttachPictureforQuote']; pt_register('POST','Howdidyoufindus'); if($FullName=="" || $Email=="" ){ $errors=1; $error.="<li>You did not enter one or more of the required fields. Please go back and try again."; } if($HTTP_POST_FILES['AttachPictureforQuote']['tmp_name']==""){ } else if(!is_uploaded_file($HTTP_POST_FILES['AttachPictureforQuote']['tmp_name'])){ $error.="<li>The file, ".$HTTP_POST_FILES['AttachPictureforQuote']['name'].", was not uploaded!"; $errors=1; } if(!eregi("^[a-z0-9]+([_\\.-][a-z0-9]+)*" ."@"."([a-z0-9]+([\.-][a-z0-9]+)*)+"."\\.[a-z]{2,}"."$",$Email)){ $error.="<li>Invalid email address entered"; $errors=1; } if($errors==1) echo $error; else{ $image_part = date("h_i_s")."_".$HTTP_POST_FILES['AttachPictureforQuote']['name']; $image_list[4] = $image_part; copy($HTTP_POST_FILES['AttachPictureforQuote']['tmp_name'], "files/".$image_part); $where_form_is="http".($HTTP_SERVER_VARS["HTTPS"]=="on"?"s":"")."://".$SERVER_NAME.strrev(strstr(strrev($PHP_SELF),"/")); $message="Full Name: ".$FullName." Email: ".$Email." Phone: ".$Phone." Questions or Comments: ".$QuestionsorComments." Attach Picture for Quote: ".$where_form_is."files/".$image_list[4]." How did you find us: ".$Howdidyoufindus." "; $message = stripslashes($message); mail("webmaster@gmail.com","Form Submitted at your website",$message,"From: phpFormGenerator"); ?> <!-- This is the content of the Thank you page, be careful while changing it --> <h2>Thank you!</h2> <table width=50%> <tr><td>Full Name: </td><td> <?php echo $FullName; ?> </td></tr> <tr><td>Email: </td><td> <?php echo $Email; ?> </td></tr> <tr><td>Phone: </td><td> <?php echo $Phone; ?> </td></tr> <tr><td>Questions or Comments: </td><td> <?php echo $QuestionsorComments; ?> </td></tr> <tr><td>Attach Picture for Quote: </td><td> <?php echo $AttachPictureforQuote; ?> </td></tr> <tr><td>How did you find us: </td><td> <?php echo $Howdidyoufindus; ?> </td></tr> </table> <!-- Do not change anything below this line --> <?php } ?> Thanks for your time and any help you can provide! I came across the PHP clone method. Does this clone first level fields and methods or does it also do deep copying e.g. copying references to those fields. For example, a field may be an array. Does it make a copy of the actual array or does it refer to the same array position in memory, so if the original changes, then the copy's changes as well?
Hello everyone, so I made this script that creates directories via "a hack around" cPanel, on my own server, in order to create directories via my own software, but I also need to copy files from one directory into that newly created directory... Therein lies the problem... I created a script a few years ago that worked, but for the life of me I can't remember how I enabled myself to be able to use the copy function of php... Anybody know? hey people i cant get copy() to work. im working on an online server and trying to copy a file into directories i just created. i read about fopen() having to be enabled if both source and destination files are urls, and fopen is enabled and both files are urls (and on the same server). the directories get created but the file doesnt get copied. the directories have permission 777. could it be that when i call the copy function the directories aren't yet available on the server? i am copying the file in the same code block as 'create folders', right after. any pointers? Hello freaks, I am currently creating a member system that will place a default handle image into a newly registered members file. I have used the mkdir function create the directory successfully based on the members ID, but I can't get the copy function to work. Code: Code: [Select] mkdir("memberFiles/$memberID", 0755); $old = 'images/defaultlogo.jpg'; $new = 'memberFiles/$memberID/logo.jpg'; copy($old, $new); Error Message: Warning: copy(memberFiles/$memberID/logo.jpg) [function.copy]: failed to open stream: No such file or directory Lastly, this folder is created to store members files. hey, can some one provide a code to help me copy a folder called contents with files in it to another folder called items cheers Hello, im new to php and I would appreciate help with this one.... Error code is Warning: copy() expects parameter 1 to be string, array given in /path/to/file/http.class.php on line 143 Error block Code: [Select] if (count($_FILES) > 0) { foreach ($_FILES as $name => $file) { if ($file['tmp_name']) { $newfile=dirname(__FILE__).'/../cache/'.$file['name']; $newfiles[]=$newfile; copy ($file['tmp_name'],$newfile); if ($file['tmp_name']) $this->post[$name]='@'.$newfile; } } } Line 143 Code: [Select] copy ($file['tmp_name'],$newfile); So my hosting server does not allow fopen anymore, so I was wondering if anyone had an alternative to copying a file from a remote server via a URL to a local folder? Any alternatives is apprecited. I have a DB with 10 fields Records are entered under a date There can be 1 to 30 records for a given date I need to copy all records of a given date with the 10 fields of each... Into the same DB only changing the original date to a new date Sounds easy, but so far I haven't been able to make it work. Any nudge in the right directiion would be much appreciated. I know this code works
INSERT INTO archive_table SELECT * FROM original_table WHERE id = 1But i want to also add the current date whenever i copy the data into a new table. Thanks in advance Hello, I need help. I want to know how to copy a field value of a table from one database to other database using php. I have a field name "invites" in a table "members" in database "db1" and a field name "new_invites" in a table "new_members" in database "db2". I want to copy value of field "invites" to "new_invites". How to do that? (Posted this in the MySQL section, but i think the php section might be the proper place)
I have a webshop db where i want to export some things linke order information to another external db on another server etc.
Example of things to move:
Table1 - orderid - orderdate Table2 - adress - shippingmethodI ONLY want it to export and import things that is not currently there. How can this be done in a php script? i am trying io copy my template to the newly created file using copy function but the copy function does not work on the server it worked fine in localhost.. i tried using fil_get_contents and file_put_function but that didnt work as well mayb cuz of the restrictions of the server.. and i do not have the permission to edit the php.ini file.. i am talking to talk with the hosting company.. but just wanted to knw is der any oder wat to copy my template to a new file that is created? Hi, I'm working on zend frame and im beginner. I have to copy text from textarea through javascript and when i press ctrl + v it should paste. And im working in linux machine. Looking for help. Please respond as soon as possible. Thanks in advance Shastry I am trying from several hours to sort and issue but i donno what the problem.. this is the following code i am using mkdir("news_upload/$year/$month/$day",0777); touch("news_upload/$year/$month/$day/$title.php"); chmod("news_upload/$year/$month/$day/$title.php",0777); $template = "news_template.php"; $new = "news_upload/$year/$month/$day/$title.php"; copy($template,$new); I have created the directories and given the permission the made a new empty file.. i added permissions to the new made file.. but it didnt apply as i check it.. (PROBLEM) then i have a template designed with the name news_template.php wanted to copy this template to the new created file.. the file gets created.. but the new file is not copying the template file on itself.. have i done anything rong??? help please? |
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