here I have value stored in $value1 variable and I want to store it's value in valuebxb.txt, and again I want to read that value from valuebxb.txt and store it to another variable $value2, here is a code from which I was able to create required file, but not able to store the value.
$value1 = round($value1, 4);
if (!file_exists('valuebxb.txt')) {
touch('valuebxb.txt', strtotime('-1 days')); //file is created
$file = fopen("valuebxb.txt","w"); //opening the file
fwrite($file,$value1); //storing the value of variable in file
fclose($file); //closing the file
}
I would recommend using file_get_contents() and file_put_contents() instead. They are much easier to use for reading and writing to files.
In your case, to write to a file:
file_put_contents('/path/to/file', $theValue);
(this will also create the file if it doesn't already exist)
And to read the value from the file:
// Define the variable with a default value
$value = null;
// Read from the file if it exists
if (is_file('/path/to/file')) {
$value = file_get_contents('/path/to/file');
}
A shorter version of reading using ternary:
$value = is_file('/path/to/file') ? file_get_contents('/path/to/file') : null;
Related
I get a ressource stream when I use fopen on a filepath I get from my database :
<?php
$query = "SELECT filepath
FROM files
WHERE id=123";
$result = do_query($query);
/*do_query() is a personal simplification*/
$line = mysql_fetch_array($result);
$file = fopen($line['filepath'], 'r');
/*var_dump on file return "resource(51) of type (stream)"*/
?>
But I my project I need a $_FILE type... Can I convert a ressource stream in $_FILE type?
$_FILES is not a "type", it's merely an array which holds certain information about uploaded files. First of all, if you have some piece of code which is hardcoded to use $_FILES, you should probably change it to accept a generic argument instead. Meaning, instead of:
function foo() {
echo $_FILES['tmp_name']:
}
Rewrite that to:
function foo($path) {
echo $path;
}
You can then call that function and pass it any path from anywhere. In your case you'd pass it $line['filepath'] as is. You don't need to fopen it, because then you get a resource, when you currently just want a path.
If you need to "fake" the $_FILES array, you need to construct it manually:
$_FILES['foo']['name'] = '..';
$_FILES['foo']['type'] = '..';
$_FILES['foo']['size'] = '..';
$_FILES['foo']['tmp_name'] = $line['filepath'];
$_FILES['foo']['error'] = UPLOAD_ERR_OK;
But again, you'll probably want to alter whatever code is hardcoded to use $_FILES instead of this hackaround.
so the title is not full clear, my question , I'm using the code to rename the file from directory present in the server the problem is i have to use the HTML form and php to update the file name, i want to do this : there will be an option on every file for renaming it when i click on the option the box pops up and i have to type the new name for file and save it , any help will be appreciated. (before down voting think about the question.)
The code that I'm using to update the file name
<?php
include("configuration.php");
$target = $_POST['filename'];
$newName = $_POST['newfilename'];
$actfoler = $_REQUEST['folder'];
$file = "files/users/";
$new ="files/users/";
$renameResult = rename($file, $new);
// Evaluate the value returned from the function if needed
if ($renameResult == true) {
echo $file . " is now named " . $new;
} else {
echo "Could not rename that file";
}
header("Location:".$_SERVER["HTTP_REFERER"]);
?>
Try changing these lines:
$file = "uploads/$loggedInUser->username$actfolder/$target";
$new ="uploads/$loggedInUser->username$actfolder/$newName";
To:
$file = "uploads/{$loggedInUser->username}{$actfolder}/{$target}";
$new ="uploads/{$loggedInUser->username}{$actfolder}/{$newName}";
To explain why:
You are using variables inside a string, which means you will want to tell PHP where the variable ends. Especially when referencing objects or arrays, but also when you are placing variables right next to each other. I'm guessing PHP evaluated your original line to uploads/[Object]->usernamePizza/newname
I don't think you can call object properties in a string as you do.
try replace these lines :
$file = "uploads/".$loggedInUser->username."$actfolder/$target";
$new ="uploads/".$loggedInUser->username."$actfolder/$newName";
You may think about echoing $file and $new to confirm the path is nicely built.
On a side note, I'd recommend to really check the entries because this code can obviously lead to major security issues.
I have a standard config file: $variable = 'value';, but at the last moment came up to use the web interface to configure it. So what is the best way to read the file, find the value of variables and then resave the file again?
At the moment I have 2 ideas:
1) RegExp
2) Keep somewhere array example
Store all config values in an associative array like so:
$config = array(
'variable' => 'value'
);
For the web interface, you can easily loop over the entire array:
foreach($config as $key=>$value) { ... }
After making changes, loop over the array and write it back to the file. (You really should be using a DB for this, though).
When you include the file, you can either use it like this:
include('config.php');
echo $config['variable']
// or
extract($config);
echo $variable;
Note: If you extract, it will overwrite any variables by the same name you might have defined before extracting.
PS - To make it easier to read and write to and from a file, I would just use json encoding to serialize the array.
Use a db
If your config is user defined - it would be better to store the config in a database. Otherwise you have this "novel" problem to solve but also potentially introduce security problems. I.e. for any one user to be able to edit your config files - they must be writeable to the webserver user. That opens the door to injecting malicious code into this file from a web exploit - or simply someone with direct access to your server (shared host?) finding this writeable file and updating it to their liking (e.g. putting "<?php header('Location: my site'); die;" in it).
One config variable
If you must manage it with a config file, include the file to read it, and var_export the variables to write it. That's easiest to do if there is only one config variable, that is an array. e.g.:
function writeConfig($config = array()) {
$arrayAsString = var_export($config, true);
$string = "<?php\n";
$string .= "\$config = $arrayAsString;\n";
file_put_contents('config.php', $string);
}
Allow partial updates
If you are changing only some variables - just include the config file before rewriting it:
function writeConfig($edits = array()) {
require 'config.php';
$edits += $config;
$arrayAsString = var_export($edits, true);
$string = "<?php\n";
$string .= "\$config = $arrayAsString;\n";
file_put_contents('config.php', $string);
}
Multiple config variables
If you have more than one variable in your config file, make use of get defined vars and loop on them to write the file back:
function writeConfig($_name = '', $_value) {
require 'config.php';
$$_name = $_value; // This is a variable variable
unset($_name, $_value);
$string = "<?php\n";
foreach(get_defined_vars() as $name => $value) {
$valueAsString = var_export($value, true);
$string .= "\$$name = $valueAsString;\n";
file_put_contents('config.php', $string);
}
}
The above code makes use of variable variables, to overwrite once of the variables in the config file. Each of these last two examples can easily be adapted to update multiple variables at the same time.
I have a project that needs to create files using the fwrite in php. What I want to do is to make it generic, I want to make each file unique and dont overwrite on the others.
I am creating a project that will record the text from a php form and save it as html, so I want to output to have generated-file1.html and generated-file2.html, etc.. Thank you.
This will give you a count of the number of html files in a given directory
$filecount = count(glob("/Path/to/your/files/*.html"));
and then your new filename will be something like:
$generated_file_name = "generated-file".($filecount+1).".html";
and then fwrite using $generated_file_name
Although I've had to do a similar thing recently and used uniq instead. Like this:
$generated_file_name = md5(uniqid(mt_rand(), true)).".html";
I would suggest using the time as the first part of the filename (as that should then result in files being listed in chronological/alphabetic order, and then borrow from #TomcatExodus to improve the chances of the filename being unique (incase of two submissions being simultaneous).
<?php
$data = $_POST;
$md5 = md5( $data );
$time = time();
$filename_prefix = 'generated_file';
$filename_extn = 'htm';
$filename = $filename_prefix.'-'.$time.'-'.$md5.'.'.$filename_extn;
if( file_exists( $filename ) ){
# EXTREMELY UNLIKELY, unless two forms with the same content and at the same time are submitted
$filename = $filename_prefix.'-'.$time.'-'.$md5.'-'.uniqid().'.'.$filename_extn;
# IMPROBABLE that this will clash now...
}
if( file_exists( $filename ) ){
# Handle the Error Condition
}else{
file_put_contents( $filename , 'Whatever the File Content Should Be...' );
}
This would produce filenames like:
generated_file-1300080525-46ea0d5b246d2841744c26f72a86fc29.htm
generated_file-1300092315-5d350416626ab6bd2868aa84fe10f70c.htm
generated_file-1300109456-77eae508ae79df1ba5e2b2ada645e2ee.htm
If you want to make absolutely sure that you will not overwrite an existing file you could append a uniqid() to the filename. If you want it to be sequential you'll have to read existing files from your filesystem and calculate the next increment which can result in an IO overhead.
I'd go with the uniqid() method :)
If your implementation should result in unique form results every time (therefore unique files) you could hash form data into a filename, giving you unique paths, as well as the opportunity to quickly sort out duplicates;
// capture all posted form data into an array
// validate and sanitize as necessary
$data = $_POST;
// hash data for filename
$fname = md5(serialize($data));
$fpath = 'path/to/dir/' . $fname . '.html';
if(!file_exists($fpath)){
//write data to $fpath
}
Do something like this:
$i = 0;
while (file_exists("file-".$i.".html")) {
$i++;
}
$file = fopen("file-".$i.".html");
Is it possible to write at a particular location in a CSV file using PHP?
I don't want to append data at the end of the CSV file. But I want to add data at the end of a row already having values in the CSV.
thanks in advance
No, it s not possible to insert new data in the middle of a file, due to filesystem nature.
Only append at the end is possible.
So, the only solution is to make another file, write a beginning part of source, append a new value, and then append the rest of the source file. And finally rename a resulting file to original name.
There you go. Complete working code:
<?php
//A helping function to insert data at any position in array.
function array_insert($array, $pos, $val)
{
$array2 = array_splice($array, $pos);
$array[] = $val;
$array = array_merge($array, $array2);
return $array;
}
//What and where you want to insert
$DataToInsert = '11,Shamit,Male';
$PositionToInsert = 3;
//Full path & Name of the CSV File
$FileName = 'data.csv';
//Read the file and get is as a array of lines.
$arrLines = file($FileName);
//Insert data into this array.
$Result = array_insert($arrLines, $PositionToInsert, $DataToInsert);
//Convert result array to string.
$ResultStr = implode("\n", $Result);
//Write to the file.
file_put_contents($FileName, $ResultStr);
?>
Technically Col. Shrapnel's answer is absolutely right.
Your problem is that you don't want to deal with all these file operations just to change some data. I agree with you. But you're looking for the solution in a wrong level. Put this problem in a higher level. Create a model that represents an entity in your CSV database. Modify the model's state and call its save() method. The method should be responsible to write your model's state in CSV format.
Still, you can use a CSV library that abstracts low level operations for you. For instance, parsecsv-for-php allows you to target a specific cell:
$csv = new parseCSV();
$csv->sort_by = 'id';
$csv->parse('data.csv');
# "4" is the value of the "id" column of the CSV row
$csv->data[4]['firstname'] = 'John';
$csv->save();