I'm just starting out writing this code and when I add in the db select and refresh the page, instead of showing all the other html on the page or an error, it just shows up blank.
Here's what I've got-
$link = mysql_connect('vps2.foo.com:3306', 'remote_vhost30', 'password');
if (!$link) {
die('Could not connect: ' . mysql_error());
}
$db_selected = mysql_select_db('best_of', $link);
if (!$db_selected) {
die ('Can\'t use foo : ' . mysql_error());
}
$sections_result = "SELECT * FROM sections";
$sections_query = mysql_query($sections_result) or die(mysql_error());
$sections_array = mysql_fetch_array($sections_result) or die(mysql_error());
This code above returns a blank page. If I comment out the row starting with $db_selected the page loads fine. Obviously it doesn't do anything with the data but no errors.
What's the problem? (And yes, I am connecting to a remote server, but the $link produces no errors)
The last line of code should be:
$sections_array = mysql_fetch_array($sections_query) or die(mysql_error());
You are trying to fetch rows from the variable $sections_result, which is your query string and not the result set.
Turn on error reporting, with error_reporting(E_ALL) like mentioned in one of the other answers.
Incidentally, I suspect the problem is that PHP is throwing an error, but you've disabled the display of errors - hence the display of a blank white page. Check the status of 'display_errors' within your php.ini file.
NB: If this is a production server, you should leave display_errors set to off.
Check it really is that line by replacing this:
$db_selected = mysql_select_db('best_of', $link);
With this:
if (! $db_selected = mysql_select_db('best_of', $link)) die('Unable to select database');
As MitMaro says you've muddled _result and _query. This might be better:
$sections_query = "SELECT * FROM sections";
$sections_result = mysql_query($sections_query) or die(mysql_error());
$sections_array = mysql_fetch_array($sections_result) or die(mysql_error());
Hope that helps :)
Related
I have tried a ton of different versions of this code, from tons of different websites. I am entirely confused why this isn't working. Even copy and pasted code wont work. I am fairly new to PHP and MySQL, but have done a decent amount of HTML, CSS, and JS so I am not super new to code in general, but I am still a beginner
Here is what I have. I am trying to fetch data from a database to compare it to user entered data from the last page (essentially a login thing). I haven't even gotten to the comparison part yet because I can't fetch information, all I am getting is a 500 error code in chrome's debug window. I am completely clueless on this because everything I have read says this should be completely fine.
I'm completely worn out from this, it's been frustrating me to no end. Hopefully someone here can help. For the record, it connects just fine, its the minute I try to use the $sql variable that everything falls apart. I'm doing this on Godaddy hosting, if that means anything.
<?php
$servername = "localhost";
$username = "joemama198";
$pass = "Password";
$dbname = "EmployeeTimesheet";
// Create connection
$conn = mysqli_conect($servername, $username, $pass, $dbname);
// Check connection
if (mysqli_connect_errno) {
echo "Failed to connect to MySQL: " . mysqi_connect_error();
}
$sql = 'SELECT Name FROM Employee List';
$result = mysqli_query($conn, $sql);
if (mysqli_num_rows($result) > 0) {
while($row = mysqli_fetch_assoc($result)) {
echo "Name: " . $row["Name"]. "<br>";
}
} else {
echo "0 results";
}
mysqli_close($conn);
?>
There be trouble here:
// Create connection
$conn = mysqli_conect($servername, $username, $pass, $dbname);
// Check connection
if (mysqli_connect_errno) {
echo "Failed to connect to MySQL: " . mysqi_connect_error();
}
There are three problems here:
mysqli_conect() instead of mysqli_connect() (note the double n in connect)
mysqli_connect_errno should be a function: mysqli_connect_errno()
mysqi_connect_error() instead of mysqli_connect_error() (note the l in mysqli)
The reason you're getting a 500 error is that you do not have debugging enabled. Please add the following to the very top of your script:
ini_set('display_errors', 'on');
error_reporting(E_ALL);
That should prevent a not-so-useful 500 error from appearing, and should instead show the actual reason for any other errors.
There might be a problem here:
$result = mysqli_query($conn, $sql);
if (mysqli_num_rows($result) > 0) {
If the query fails, $result will be false and you will get an error on the mysqli_num_rows() call. You should add a check between there:
$result = mysqli_query($conn, $sql);
if (!$result) {
die('Query failed because: ' . mysqli_error($conn));
}
if (mysqli_num_rows($result) > 0) {
The name of your database table in your select statement has a space in it. If that is intended try:
$sql = 'SELECT Name FROM `Employee List`';
i think you left blank space in your query.
$sql = 'SELECT Name FROM Employee List';
change to
$sql = "SELECT `Name` FROM `EmployeeList`";
I'm trying to make a simple PHP script that fetches a table from my MySQL database and encodes the results in JSON, so I can use them later in Java.
This is my code:
<?php
$servername = "localhost:3036";
$username = "example_user";
$password = "example_password";
$conn = mysql_connect($servername, $username, $password);
if(! $conn) {
die("Could not connect: " . mysql_error());
}
$sql = "SELECT * FROM table_name";
mysql_select_db("database_name");
$retval = mysql_query($sql, $conn);
if(! $retval) {
die("Could not get data: " . mysql_error());
}
while($row = mysql_fetch_assoc($retval)) {
$output[]=$row;
}
print(json_encode($output));
mysql_close($conn);
?>
This just gives a blank page as output (error messages are set to display).
However, if I change json_encode($output) to json_encode($output[0]) (or any other number within the array's bounds), the output becomes that one $row array.
This is probably a really stupid question, but after about 3 hours of research I'm at my wit's end. Thank you for any help.
User #Joni led me to the solution.
Adding mysql_set_charset("utf8") fixed my issue.
As mentioned in this post: Why is this PHP call to json_encode silently failing - inability to handle single quotes?.
Try
echo json_encode($output) ;
It seems you have some utf8 character in your result set
add this statement before running your query
mysql_query('SET CHARACTER SET utf8');
Update
"mysql"
to
"mysqli"
and add
mysqli_set_charset($variável_de _conexão, 'utf8');
below the connection variable
Im using phpgraph lib to create graphs on my linux server. I tried an example and it worked, but I had provided it with the data.
then I wanted to connect it to mysql database and plot a query, when I run it, nothing happens, I don't see any output on the page or any errors, I don't see any output on the page at all, even if I put wrong credentials to my database e.t.c any inputs?
I have executed the sql statement on sql server and it's working fine.
the version of php the server has is PHP 5.3.3
<?php
include('phpgraphlib.php');
$graph= new PHPGraphLib(550,350);
$link = mysql_connect('localhost', 'user', 'password')
or die('Could not connect: ' . mysql_error());
mysql_select_db('databasename' or die('Could not select database');
$dataArray=array();
//get data from database
$sql="my sql statement";
$result = mysql_query($sql) or die('Query failed: ' . mysql_error());
if ($result) {
while ($row = mysql_fetch_assoc($result)) {
$salesgroup=$row["var1"];
$count=$row["count"];
//add to data areray
$dataArray[$salesgroup]=$count;
}
}
//configure graph
$graph->addData($dataArray);
$graph->setTitle("Sales by Group");
$graph->setGradient("lime", "green");
$graph->setBarOutlineColor("black");
$graph->createGraph();
?>
I fixed it, I was expecting to see errors on the webpage, but didn't see any on CHROME, I hen opened it in IE and saw error 500.
Troubleshooted through the log file.
Turned out the sql statement wasn't suppose to have double quotes e.g instead of
where name="john"
it's suppose to be
where name='john'
Got a problem! Though I found almost similar threads but none helped :(
I've written a php script to fetch the number of registered users from my MySQL database. The script is working great in my localhost; it is using the given username,pass and host name which are "root", "root", and "localhost" respectively, but the script is not using the given username/pass/host rather using root#localhost (password: NO) in Live server.
In the Live server I created a MySQL user, set an different password, and hostname there is of course not localhost. I updated the script with my newly created mysql users data. BUT, whenever I run the script, I see that the script is still using "root", "root", and "localhost"!!
take a look at the script:
//database connection
$conn = mysql_connect( "mysql.examplehost.com", "myusername", "mypass" );
$db = mysql_select_db ("regdb",$conn); //Oops, actually it was written this way in the script. I misstyped it previously. now edited as it is in the script.
//Query to fetch data
$query = mysql_query("SELECT * FROM regd ");
while ($row = mysql_fetch_array($query)):
$total_regd = $row['total_regd'];
endwhile;
echo $total_regd;
-- Some says to change the default username and pass in the config.ini.php file located in phpMyAdmin directory. Would this help?? I didn't try this because either my hosting provider didn't give me privilege to access that directory (because I am using free hosting for testing scripts) or I simply didn't find it :(
Please help....
Foreword: The MySQL extension is marked as deprecated, better use mysqli or PDO
Though you store the connection resource in $conn you're not using it in your call to mysql_query() and you're not checking the return value of mysql_connect(), i.e. if the connection fails for some reason mysql_query() "is free" to establish a new default connection.
<?php
//database connection
$conn = mysql_connect( "mysql.examplehost.com", "myusername", "mypass" );
if ( !$conn ) {
die(mysql_error()); // or a more sophisticated error handling....
}
$db = mysql_select_db ("regdb", $conn);
if ( !$db ) {
die(mysql_error($conn)); // or a more sophisticated error handling....
}
//Query to fetch data
$query = mysql_query("SELECT * FROM regd ", $conn);
if (!$query) {
die(mysql_error($conn)); // or a more sophisticated error handling....
}
while ( false!=($row=mysql_fetch_array($query)) ):
$total_regd = $row['total_regd'];
endwhile;
echo $total_regd;
edit: It looks like you're processing only one row.
Either move the echo line into the while-loop or (if you really only want one record) better say so in the sql statement and get rid of the loop, e.g.
// Query to fetch data
// make it "easier" for the MySQL server by limiting the result set to one record
$query = mysql_query("SELECT * FROM regd LIMIT 1", $conn);
if (!$query) {
die(mysql_error($conn)); // or a more sophisticated error handling....
}
// fetch data and output
$row=mysql_fetch_array($query);
if ( !$row ) {
echo 'no record found';
}
else {
echo htmlspecialchars($row['total_regd']);
}
First of all:
$link = mysql_connect('localhost', 'mysql_user', 'mysql_password');
if (!$link) {
die('Not connected : ' . mysql_error());
}
// make foo the current db
$db_selected = mysql_select_db('foo', $link);
if (!$db_selected) {
die ('Can\'t use foo : ' . mysql_error());
}
What is your mysql_error()? :)
I have a problem with my php script through which i need to send data to db. When i send data, i get blank screen, no matter what I write, select and everything else.
<?php
$con = mysql_connect("localhost","root","");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db("gagq", $con);
$result = mysql_query("SELECT * FROM gagq.user WHERE id=1;");
echo $result;
mysql_close($con);
?>
basically , i get blank screen from this. i access the file through the form, submit button. Method post, action name of file. Now, i've filled the db on id=1, but no matter what i write, on php nothing works. Any idea?
try this
$result = mysql_query("SELECT * FROM gagq.user WHERE id=1;");
$array = mysql_fetch_assoc($result);
var_dump($array);
When you are in a testing phase, you should include:
/* Turns Error Reporting ON */
ini_set('display_errors',1);
error_reporting(E_ALL);
in order to see and correct any php errors, rather than adding die checking everywhere.