I just set up the prepared query selection but it seems like it does not work.. It shows blank page instead of the content it should show..where is problem?
<?php
$servername = "";
$username = "";
$password = "";
$dbname = "";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$stmt = $mysqli->prepare('SELECT configured FROM members WHERE username = ?');
$stmt->bind_param('s', $username);
$stmt->execute();
$stmt->bind_result($username);
while ($stmt->fetch()) {
if($row['configured'] == ""){
if (isset($_SESSION['user_id'])) {
?>
With this code I want to check if user has configured his settings and so if yes, then if he is logged in. But the page shows nothing even though I have there everything correct after that code I posted...
You're using $conn to connect with, then $mysqli in your prepare statement.
This
$stmt = $mysqli->prepare
Which should read as
$stmt = $conn->prepare
Also make sure you've started the session.
http://php.net/manual/en/function.session-start.php
Check for errors:
http://php.net/manual/en/function.error-reporting.php
and you have 3 missing closing braces in your posted code.
I see you're using $username = ""; in your DB connection and then using the same variable to bind with.
Add error reporting to the top of your file(s) which will help find errors.
<?php
error_reporting(E_ALL);
ini_set('display_errors', 1);
// rest of your code
Sidenote: Error reporting should only be done in staging, and never production.
Related
I can't Update my Database with PHP. I don't get any errors but it doesn't change anything!
Here is my file:
<?php
include_once 'dbh.inc.php';
?>
<?php
$id = $_GET['verId'];
$name = $_GET['verName'];
echo $id;
echo $name;
$sql = "UPDATE allusers SET ver = '1' WHERE idUsers = '$id';";
?>
The variables are defined and work.
Here's the dbh.inc.php file:
<?php
$servername = "localhost";
$dBUsername = "root";
$dBPassword = "";
$dBName = "loginsystem";
$conn = mysqli_connect($servername, $dBUsername, $dBPassword, $dBName);
if (!$conn) {
die("Connection failed: ".msqli_connect_error());
}
?>
Other files that use dbh.inc.php work fine.
Thanks for your help.
You need to execute your SQL, but the way you are using MySQLi is very wrong. Let me show you how to get started with a simple query.
First in your dbh.inc.php (you should name it properly too) you should have the following code:
<?php
mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);
$conn = new \mysqli("localhost", "root", "", "loginsystem");
$conn->set_charset('utf8mb4');
Do not use root for connection. Create a valid MySQL user with a proper password.
Then in your main PHP file, you can use it as follows:
<?php
include_once 'dbh.inc.php';
$id = $_GET['verId'];
$name = $_GET['verName'];
echo $id;
echo $name;
// prepare -> bind data -> execute
$stmt = $conn->prepare("UPDATE allusers SET ver='1' WHERE idUsers=?");
$stmt->bind_param('s', $id);
$stmt->execute();
I used here what is called a prepared statement. You can learn more about MySQLi here: https://phpdelusions.net/mysqli_examples/update
I have looked at several examples on how to call a MySQL stored procedure from PHP but none have helped me. The stored procedure works when run inside PHPMyAdmin but I am having trouble calling it from the web.
<?php
$servername = "localhost";
$username = "username";
$password = "password";
$dbname = "dbname";
$conn = new mysqli($servername, $username, $password, $dbname);
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$result = mysqli_query($conn,"CALL standings_build()");
if (mysqli_query($conn,$sql))
header('refresh:1; url=schedule_main_scores.php');
else
echo "failed";
?>
There's 2 problems here.
You're querying twice and using the wrong variable, being $sql instead of $result.
$result = mysqli_query($conn,"CALL standings_build()");
if (mysqli_query($conn,$sql))
^^^^^^^^^^^^ calling the query twice
^^^^ wrong variable, undefined
all that needs to be done is this:
if ($result)
and an else to handle the (possible) errors.
Error reporting and mysqli_error($conn) would have been your true friends.
http://php.net/manual/en/function.error-reporting.php
http://php.net/mysqli_error
Side note: You really should use proper bracing techniques though, such as:
if ($result){
echo "Success";
}
else {
echo "The query failed because of: " . mysqli_error($conn);
}
It helps during coding also and with an editor for pair matching.
I have created one HTML form which takes input from user ,Now I need to search user inputed name in Mysql database and print details related to that user inputed name which is stored in Mysql database.
Below script is creating HTML form to take user input, Saved as "ProcessTracking.html".
<form action="details.php" method="get"/>
<h3 align="center"><FONT color=#CCFF66>ENTER SO NUMBER</h3>
<p align="center">
<input type="text" id="SO_Number" name="SO_Number"/>
</p>
<div style="text-align:center">
<button type="submit" value="SEARCH">
<img alt="ok" src=
"http://www.blueprintcss.org/blueprint/plugins/buttons/icons/tick.png"/>
SEARCH
</button>
</form>
Below PHP script named as "details.php"
<?php
$userinput = $_GET['SO_Number'];
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "ProcessTrackingSystem";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_errno) {
printf("Connect failed: %s\n", $conn->connect_error);
exit();
}
$result = mysqli_query($conn, "SELECT * FROM ProcessTrackingSystem.ProcessDetails WHERE SO_Number = '$userinput'") or die(mysqli_error($conn));
$row = mysqli_fetch_assoc($result);
#printf ("SO_Number: %s \n",$row["SO_Number"])
#print_r($row);
printf ("SO_Number:");
printf($row["SO_Number"]);
printf ('--||--');
printf ("Name:");
printf($row["Name"]);
printf ('--||--');
$conn->close();
?>
Firstlly you are not using the $_GET['SO_Number'] parameter in a WHERE of SQL statement. Secondlly you are using both mysql and mysqli which are totaly diffrent and don't work together. For usage see mysqli_fetch_row() and mysqli_query(). Also use print_r($row);.
Here is the corrected code:
<?php
$userinput = $_GET['SO_Number'];
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "ProcessTrackingSystem";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_errno) {
printf("Connect failed: %s\n", $conn->connect_error);
exit();
}
$result = mysqli_query($conn, "SELECT * FROM ProcessTrackingSystem.ProcessDetails WHERE SO_Number = '$userinput'") or die(mysqli_error($conn));
$row = mysqli_fetch_row($result);
print_r($row);
$conn->close();
?>
EDIT: Added code example.
You have mixed mysql and mysqli api together. Try using either one.
Note: mysql api is deprectaed as of php 5.5.0
1st Error
As saty says it is because of the syntax error you have in this line
print_r"$row";
which should be as print_r($row)
2nd Error
You're mixing mysql & mysqli
I am not sure about the table that you have.
Recommendation :
I would recommend you to turn on the error_reporting if not those errors will be in your errors_log file
Also for debugging your sql, you can first construct your sql query, run in the phpmyadmin or related tools for your query, then fire the query and make this done.
Note :
If you are using these code in online then the error_log will be in the directory where you execute this page. (But it may change according to your hosting)
If you are running in local machine the error log may locate according to the server you use...
You can find by printing the php's configuration by phpinfo and find for
error_log
May this thing will fix your issue
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "ProcessTrackingSystem";
$so = $_POST['SO_Number'];
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$stmt = $conn->prepare("SELECT * FROM ProcessDetails WHERE SO_Number=$so");
$stmt->execute();
$result = $stmt->get_result();
$row = $result->fetch_assoc();
print_r($row[SO_Number]);
$conn->close();
?>
I'm an iOS developer trying to update a bit 'o php to work with the mysqli_* stuff instead of the deprecated mysql_* stuff.
I know next to nothing about php/mysql, and I'm stuck. I'm using a script found at http://www.w3schools.com/php/php_mysql_select.asp, with changes to reflect my field names etc
When I call the following from a browser I get an access error (Connection failed: Access denied for user 'whoisit7'#'localhost' (using password: YES)). The server name, password and username etc I'm using all work with an existing script in a browser but not with this new one.
Can anyone point me at what I'm getting wrong?
<?php
$servername = "localhost";
$userName = "whoisit7_ios";
$password = "blahblahblah";
$dbName = "whoisit7_scathing";
// Create connection
$conn = mysqli_connect($servername, $username, $password, $dbname);
// Check connection
if (!$conn) {
die("Connection failed: " . mysqli_connect_error());
}
$sql = "SELECT name, latitude, longitude FROM location where status = 1";
$result = mysqli_query($conn, $sql);
if (mysqli_num_rows($result) > 0) {
// output data of each row
while($row = mysqli_fetch_assoc($result)) {
echo "name: " . $row["name"]. " - latlong: " . $row["latitude"]. " " . $row["longitude"]. "<br>";
}
} else {
echo "0 results";
}
mysqli_close($conn);
?>
In PHP, variable names are case-sensitive. So, Change your connection variable names to:
$servername = "localhost";
$username = "whoisit7_ios";
$password = "blahblahblah";
$dbname = "whoisit7_scathing";
Read it if you are new to PHP: http://php.net/manual/en/language.variables.basics.php
Php is case sensitive, so change the below line from:
$conn = mysqli_connect($servername, $username, $password, $dbname);
to
$conn = mysqli_connect($servername, $userName, $password, $dbName);
You have made error in $username which is $userName and $dbname which should be $dbName as you declared above.
Just change -->
localhost to 127.0.0.1 //you need to change params in mysql config if u want to use it as localhost
And
$username to $userName //As u need to keep these variables similar :)
this Problem seems to be coming from database privileges as i have also have encountered this before. Give this fix a try.
GRANT ALL PRIVILEGES ON database_name TO usernamehere#host IDENTIFIED BY 'yourpassword';
FLUSH PRIVILEGES;
Just change the lower case letters to your own credentials.
I have a php script I wrote and it will not return anything except "Hello".
<?php
echo "Hello";
$username = "me";
$password = "password";
$hostname = "localhost";
$dbname = "dbname";
//connection to the database
$dbhandle = new mysqli($hostname, $username, $password, $dbname)
or die("Unable to connect to MySQL");
echo "logged in";
// Sanitize variable
$user_id = intval($_GET['user_id']);
//update likes on specific id
$result = mysqli_query($dbhandle, "SELECT user_score FROM users WHERE id = '$user_id'")
or die("Unable to query");
// Send back the score
$return = array();
while ($row = mysqli_fetch_assoc($result)) {
$return[] = $row;
}
print json_encode($return);
//close the connection
mysqli_close($dbhandle);
?>
I had this script working before, then I moved it to another instance. I know I can login with my credentials to mysql and check that I have all privileges, and I know my username and password are correct. Is there something I need to change in my php? Or is there something I need to do so that when I access through a browser it shows like turn on the server?
I am not having anything returned except Hello, it does not even tell me "Unable to connect". This is very confusing to me. I should at least see "logged in" OR "Unable to connect" right?
Your are mixing the object oriented and procedural mysqli functions. When using the mysqli-constructor an object is returned even on failure, therefore the error is never shown.
Use this code instead to connect to your database:
$dbhandle = mysqli_connect($hostname, $username, $password, $dbname) or die("mysqli_connect failed");
Check the mysqli-documentation for further information.
You probably aren't getting any output because of your error_reporting settings in PHP. The first step would be to set display_errors = On in your php.ini, along with error_reporting = E_ALL. This will show all errors and you will get more information from your code snippet above. You can also set the error_reporting level in the PHP file itself by adding this line at the top of your file: error_reporting(-1);
More info on error_reporting here.