This question already has answers here:
PHP: How to check if image file exists?
(22 answers)
Closed 5 years ago.
Can someone please help me to make this script check if the file exists, before it truncate the table.
If filename not exists, I want to stop the import.
<?php
//set the connection variables
$hostname = "host";
$username = "username";
$password = "pass";
$database = "database";
$filename = "filename.csv";
//connect to mysql database
$connection = mysqli_connect($hostname, $username, $password, $database) or die("Error " . mysqli_error($connection));
mysqli_query($connection, "TRUNCATE TABLE `my_tablename`");
// open the csv file
$fp = fopen($filename,"r");
//parse the csv file row by row
while(($row = fgetcsv($fp,"500",",")) != FALSE)
{
//insert csv data into mysql table
$sql = "INSERT INTO Pristabell (Produkt, Pris, Rabattkr, Rabattprosent, Lagerstatus, Butikk, TAGS) VALUES('" . implode("','",$row) . "')";
if(!mysqli_query($connection, $sql))
{
die('Error : ' . mysqli_error($conection));
}
}
fclose($fp);
//close the db connection
mysqli_close($connection);
?>
Thanks :-)
http://php.net/manual/en/function.file-exists.php
if(file_exists($pathtofile)){
//do import
}else{
//stop
}
One simple solution is use
file_exist;
in an if() chunk.
Before the while(), if true continue else exit or trow an exception.
Related
I have a wordpress plugin that exports form entries to a txt file. So I need to write a php script to add them to a sql database as I want the submissions added to a database on a different domain (otherwise I’d just get the plugin to do it for me). I’m fine about how I get it to connect to the database, it’s just how I code it to interpret the data as the column names are always next to the field as shown.
{"Entry_ID":"235","Name":"matt","Email":"matt#gmail.com","Date":"03/10/2017"}{"Entry_ID":"236","Name":"matt","Email":"matt#btinternet.com","Date":"10/10/2017"}
Is there a way to get it to ignore the column name and only interpret the data within the “” after the : ?
Once these have been added to the sql database I would then need to get the lines removed from the txt
So far I have this but it isn't working...
$file= fopen('http://mpcreations.staging.wpengine.com/wp-content/themes/red-seal-resources/test.txt', 'r');
while (($data = fgetcsv($file)) !== FALSE) {
$object = json_encode($data[0]);
$servername = "";
$username = "";
$password = "";
$dbname = "";
// Create connection
$conn = mysqli_connect($servername, $username, $password, $dbname);
// Check connection
if (!$conn) {
die("Connection failed: " . mysqli_connect_error());
}
$query = "INSERT INTO 'wp_forms' LINES TERMINATED BY '\n';
if (mysqli_multi_query($conn, $query)) {
echo "New records created successfully";
} else {
echo "Error: " . $query . "<br>" . mysqli_error($conn);
}
mysqli_close($conn);
}
Any help would be greatly appreciated.
Thank you
Each line in the txt file has JSON data? Process the txt file, parse the data and INSERT it into the database table.
$file= fopen('file.txt', 'r');
while (($data = fgetcsv($file)) !== FALSE) {
$object = json_encode($data[0]);
// Prepare INSERT query here...
}
This question already has answers here:
Reference - What does this error mean in PHP?
(38 answers)
Closed 5 years ago.
solved: i missed ; in echo file
I am new to php, i have succesfully loaded data into sql using input textbox in html but i am unable to load the same into database by reading csv file and i am getting parse error for the fopen function. pls help
upload.php
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "ib";
$conn = new mysqli($servername, $username, $password, $dbname);
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
} else
$file=$_FILES["file"]["tmp_name"];
echo $file
fopen($file, 'r');
while (($emapData = fgetcsv($file, 10000, ",")) !== FALSE)
$sql = "INSERT into ib(atmid,location,zone,cash,fault) values('.$emapData[0]','.$emapData[1]','.$emapData[2]','.$emapData[3]','.$emapData[4])";
$conn->query($sql);
}
fclose($test);
echo "CSV File has been successfully Imported.";
?>
You $sql has quotes problem, below is the updated one:
$sql = "INSERT into ib(atmid,location,zone,cash,fault) values('".$emapData[0] ."','".$emapData[1] ."','" . $emapData[2] . "','".$emapData[3] . "','".$emapData[4] . "')";
This question already has answers here:
Can I mix MySQL APIs in PHP?
(4 answers)
Closed 5 years ago.
I want to insert data to phymyadmin table using CSV file from front-end of website, which must UPDATE all the record previously in my table.
This code isn't working at all. Please locate the error.
<?php
$servername = "localhost";
$username = "";
$password = "";
$dbname = "";
$conn = mysql_connect($servername, $username , $password );
if(! $conn ) {
die('Could not Find Some Error Occured: ' . mysql_error());
}
if(isset($_POST["submit"]))
{
if($_FILES['file']['name'])
{
$filename = explode(".", $_FILES['file']['name']);
if($filename[1] == 'csv')
{
$handle = fopen($_FILES['file']['tmp_name'], "r");
while($data = fgetcsv($handle))
{
$uname = mysqli_real_escape_string($connect, $data[0]);
$pass = mysqli_real_escape_string($connect, $data[1]);
mysql_select_db('desiresq_record');
$query = "INSERT into login (username, password)
VALUES ('$uname','$pass')";
mysqli_query($connect, $query);
}
fclose($handle);
echo "<script>alert('Import done');</script>";
}
}
}
?>
You're using both MySQL and the MySQLi extensions in your script.
I'd suggest just sticking to MySQLi, seeing as MySQL is long deprecated and shouldn't be used due to security issues.
Best of luck!
My PHP file need to save the image in a server and store the image path in a MYSQL database.
My database imageid contains table image_table as below :
create table image_table
(
ID INT not null AUTO_INCREMENT,
path varchar(256),
primary key (ID)
)
My PHP Code is as below. Image saving in the server is working fine, but throws few errors storing the image path in DB.
Error running the below PHP code :
Parse error: syntax error, unexpected '$conn' (T_VARIABLE) in C:\xampp\htdocs\appinventor\postfile.php on line 7
The PHP Code with above error:
<?PHP
$servername = "localhost";
$username = "root";
$password = "basis123";
$database = "imageid"
// Create connection
$conn = new mysqli($servername, $username, $password);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
echo "Connected successfully";
#mysqli_select_db($database) or die( "Unable to select database");
$query = "INSERT INTO image_table (path) VALUES('$_GET['filename']')");
//mysql_query($query);
//File Transfer Logic
$data = file_get_contents('php://input');
if (!(file_put_contents($_GET['filename'],$data) === FALSE)) echo "File xfer completed."; // file could be empty, though
else echo "File xfer failed.";
// $_GET['filename'] has the file name . and C:\xampp\htdocs\myapp is the file path
echo $_GET['filename']
//mysql_close();
?>
When i remove the database connection and SQLQuery in php code the connection is successful.
Removed lines of code for successful connection and log as "Connected successful"
#mysqli_select_db($database) or die( "Unable to select database");
$query = "INSERT INTO image_table (path) VALUES('$_GET['filename']')");
//mysql_query($query);
Where did i go wrong ?Any suggestions ?
You forgot semicolon after $database = "imageid"
$database = "imageid";
And you have extraneous parenthesis here:
$query = "INSERT INTO image_table (path) VALUES('$_GET['filename']')");
Try this:
<?php
$servername = "localhost";
$username = "root";
$password = "basis123";
$database = "imageid";
// Create connection
$conn = new mysqli($servername, $username, $password,$database);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
echo "Connected successfully";
mysqli_select_db($conn,$database);
$query = "INSERT INTO image_table (path) VALUES('".$_GET['filename']."')";//also this is unsafe
//mysql_query($query);
//File Transfer Logic
$data = file_get_contents('php://input');
if (!(file_put_contents($_GET['filename'],$data) === FALSE)) echo "File xfer completed."; // file could be empty, though
else echo "File xfer failed.";
// $_GET['filename'] has the file name . and C:\xampp\htdocs\myapp is the file path
echo $_GET['filename']
//mysql_close();
?>
Main problem:
$database = "imageid"
^----missing ;
$conn = new m
PHP strings+arrays 101: You cannot use quoted array keys within a double-quoted string, unless you use the {}-extended syntax:
$foo = "$arr['key']"; // bad
$foo = "$arr[key]"; // ok
$foo = "{$arr['key']}"; // ok
So:
$query = "INSERT INTO image_table (path) VALUES('$_GET['filename']')");
^--------^
is wrong, as well as being vulnerable to SQL injection attacks.
You need a semi colon after
$database = "imageid"
so it should be
$database = "imageid";
You also don't want to be putting user input directly into your SQL. http://en.wikipedia.org/wiki/SQL_injection
Right now I am inserting blob files into a database. I have read up on the update syntax for mysql I can not figure out how to modify my code to update a row with the BLOB instead of inserting a new row with the BLOB. Could someone help me with this?
Here is my code:
<?php
// Create MySQL login values and
// set them to your login information.
$username = "root";
$password = "";
$host = "localhost";
$database = "test";
$tbl_name="members";
// Make the connect to MySQL or die
// and display an error.
$link = mysql_connect($host, $username, $password);
if (!$link) {
die('Could not connect: ' . mysql_error());
}
// Select your database
mysql_select_db ($database);
// Make sure the user actually
// selected and uploaded a file
if (isset($_FILES['image']) && $_FILES['image']['size'] > 0) {
// Temporary file name stored on the server
$tmpName = $_FILES['image']['tmp_name'];
// Read the file
$fp = fopen($tmpName, 'r');
$data = fread($fp, filesize($tmpName));
$data = addslashes($data);
fclose($fp);
// Create the query and insert
// into our database.
$query = "INSERT INTO members ";
$query .= "(image) VALUES ('$data')";
$results = mysql_query($query, $link);
// Print results
print "Thank you, your file has been uploaded.";
}
else {
print "No image selected/uploaded";
}
// Close our MySQL Link
mysql_close($link);
?>
1° You need to pass a referecence for what Data you are trying to update, like the Primary Key Id From Table.
2° Update SQL should be like it
$image = mysql_real_escape_string($unsafe_image);
$id = mysql_real_escape_string($unsafe_id);
$query = "UPDATE members SET image = '$data' WHERE id_image = $id";
$results = mysql_query($query, $link);