I've problem with mysql. I've PHP script, which returned array into json data from datebase.
I've message from 'echo' about successfully connection, but my result is equals which null of array.
In result on Explorer I've:
Connected successfully
query: SELECT name,id FROM rz_DWzZ'
result:
RESULT:[]
This is this script.
$conn = mysql_connect($servername, $username, $password);
mysql_select_db($database);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
echo "Connected successfully";
$return_arr = array();
$qstring = "SELECT name,id FROM rz_DWzZ";
$result = mysql_query($qstring,$conn);
echo "<br>query: ".$qstring."<br>";
echo "<br>result: ".$result."<br>";
while ($row = mysql_fetch_assoc($result))//loop through the retrieved values
{
$row['name']=htmlentities(stripslashes($row['name']));
$row['id']=(int)$row['id'];
array_push($return_arr,$row);
}
mysql_close($conn);
echo "<br>RESULT:".json_encode($return_arr);
You didn't check for failure properly. mysql_*() functions return boolean FALSE on failure, which echo will print as a zero-length/invisible string.
You have to explicitly test for it:
$result = mysql_query(...) or die(mysql_error());
^^^^^^^^^^^^^^^^^^^^^^--method #1
if ($result === false) { // method #2
die(mysql_error());
}
And of course, you should NOT be using those functions anyways. They're obsolete/deprecated, and your code is now useless in newer versions of PHP. You should be using mysqli or PDO for any new development.
As well, you have numerous other bugs:
if ($conn->connect_error) {
the mysql_*() function library has NEVER been object-oriented. It's purely procedural, and has absolutely NO object support whatsoever. Therefore this connection test will always fail, as $conn->connect_error will always evaluate to null, which converts to boolean false as well, meaning you get a false positive for success.
Related
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've looked all over here. Please be patient as I am new to php and mysql.
I got WAMPP installed & seems to be working OK. I created a simple "test" database from phpMyAdmin and "firsttable" in that. I can do a simple connect using example from w3schools, but trying to select & display data I entered only throws back errors.
<?php
$servername = "localhost";
$username = "root";
$password = "";
// Connect
$conn = mysqli_connect($servername, $username, $password);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT reference, firstname, lastname, room FROM firsttable";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
echo "id: " . $row["reference"]. " - Name: " . $row["firstname"]. " " . $row["lastname"]. "room:" . $row["room"]. "<br>";
}
} else {
echo "0 results";
}
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$conn->close();
?>
First off, I get a parse error on line 17. The one that reads:
if ($result->num_rows > 0) {
The error says: Trying to get property of non-object.
I tried wrapping the whole php code in tags and saving it as html, but then it appeared that no row data was ever found.
I am able to use very simple code that connects successfully. I can confirm the database is in there, so is the table, and the contents I added to it.
Please, what am I doing wrong?
You need to specify the database when you connect:
$database = 'test';
$conn = mysqli_connect($servername, $username, $password, $database);
where $database is the name of your database (test in this case). MySQL doesn't know which database your table resides in without you telling it.
In addition, you should always include error checking for your database connection (you have two of these, you don't need the last one) as well as any queries. Sans this, you can check your error logs for more information when something fails.
I am working on converting some PHP code from mysql to mysqli. I have created an error and am unable to understand how to fix it. Any suggestions would be greatly appreciated.
The code looks like this:
<?php
include ("admin/includes/connect.php");
$query = "select * from posts order by 1 DESC LIMIT 0,5";
$run = mysqli_query($conn["___mysqli_ston"], $query);
while ($row=mysqli_fetch_array($run)){
$post_id = $row['post_id'];
$title = $row['post_title'];
$image = $row['post_image'];
?>
The error produced is: Fatal error: Cannot use object of type mysqli as array
The error is being called out on this line:
$run = mysqli_query($conn["___mysqli_ston"], $query);
In the line above $conn is a variable from the database connect file which has this code:
<?php
// Stored the db login credentials in separate file.
require("db_info.php");
// Supressing automated warnings which could give out clues to database user name, etc.
mysqli_report(MYSQLI_REPORT_STRICT);
// Try to open a connection to a MySQL server and catch any failure with a controlled error message.
try {
$conn=mysqli_connect ('localhost', $username, $password) or die ("$dberror1");
} catch (Exception $e ) {
echo "$dberror1";
//echo "message: " . $e->message; // Not used for live production site.
exit;
}
// Try to Set the active MySQL databaseand catch any failure with a controlled error message.
try {
$db_selected = mysqli_select_db($conn, $database) or die ("$dberror2");
} catch (Exception $e ) {
echo "$dberror2";
//echo "message: " . $e->message; // Not used for live production site.
exit;
// We want to stop supressing automated warnings after the database connection is completed.
mysqli_report(MYSQLI_REPORT_OFF);
}
?>
This line
$run = mysqli_query($conn["___mysqli_ston"], $query);
should be
$run = mysqli_query($conn, $query);
If you're migrating to mysqli, you should really read these docs at least.
The proper way to use a mysqli connection:
<?php
$mysqli = new mysqli("example.com", "user", "password", "database");
if ($mysqli->connect_errno) {
echo "Failed to connect to MySQL: (" . $mysqli->connect_errno . ") " . $mysqli->connect_error;
}
$res = $mysqli->query("SELECT id FROM test ORDER BY id ASC");
while ($row = $res->fetch_assoc()) {
echo " id = " . $row['id'] . "\n";
}
?>
you should also consider utilizing mysqli's prepared statements
I have connected to a database for the first time with oop and stright away come up with an issue, below is my code which i'm struggling with:
$q = 'SELECT * FROM test';
$sqli->query($q);
if($sqli->query($q)){
echo "worked";
}
if($sqli->error){
echo $sqli->error;
}
I have checked for errors when connecting to the db and that works fine, but when I run this query I get no output, why? I expected an error or "worked", but have got neither.
Whats happening?
I have put some comments in the source code to help:
$q = 'SELECT * FROM test';
//$sqli is the result of a
//new mysqli("localhost", "user", "password", "database");
$resource = $sqli->query($q); // this returns a resource or false
if(!$resource) {
echo $sqli->error;
die; // do not process further
}
// process the results
$rows = $resource->fetch_all();
if ($rows) { // check if there are rows
echo "worked";
}
else {
echo "query is ok, but there are no rows";
}
You could also use $resource->fetch_object() which returns an object for output. Therefore if you wanted to print specific data from the result set, you would do something like
//table test.Name and test.Country
while ($rowobj = $resource->fetch_object()){
printf ("%s (%s)\n", $rowobj->Name, $rowobj->Country);
}
Good luck,
You could use this method, I hope it's what you are looking for. You will need to define the DB first. Then you can connect in OOP and test the connection is true or exit();
Let me know if this works for you. You can also define the DB in an external file and just do an include(); towards the top of your script for any pages needing connection to the DB.
define("SERVER","IP Address");
define("USER","DB USERNAME");
define("PASSWORD","DB PASSWORD");
define("DATABASE","DB NAME");
// This is for connection
$mysqli = new mysqli(SERVER, USER, PASSWORD, DATABASE);
if ($mysqli->connect_errno) {
echo "Connection to MySQL failed: (" . $mysqli->connect_errno . ") " . $mysqli->connect_error;
exit();
}
I am using this code to insert some values in MySql table:
<?php
mysql_connect("localhost","root","root");
mysql_select_db("bib");
$id = "12";
$titlu = "Joe";
$query = "INSERT INTO carte SET id='$id', titlu='$titlu'";
$result = mysql_query($query);
// Display an appropriate message
if ($result)
echo "<p>Product successfully inserted!</p>";
else
echo "<p>There was a problem inserting the Book!</p>";
mysql_close();
?>
After running it into browser, the following error occurs:
"Apache HTTP Server has encountered a problem and needs to close. We are sorry for the inconvenience."
It seems that mysql_select_db("bib") statement causes it. Database is create , also table...
I am running php 5.3 and mysql 5.1 on windows xp sp 2.
Please any ideas are welcomed...
Thanks...
Any of the mysql_* functions can fail for various reasons. You have to check the return values and if a function indicates an error (usually by returning FALSE) your script has to react appropriately.
mysql_error($link) and mysql_errno($link) can give you more detailed information about the cause. But you don't want to show all the details to just any arbitrary user, see CWE-209: Information Exposure Through an Error Message.
If you don't pass the connection resource returned by mysql_connect() to subsequent mysql_* functions calls, php assumes the last successfully established connection. You shouldn't rely on that; better pass the link resource to the functions. a) If you ever have more than one connection per page you must pass it anyway. b) If there is no valid db connection the php-mysql modules tries to establish the default connection which is usually not what you want; it only takes up more time to fail ..again.
<?php
define('DEBUGOUTPUT', 1);
$mysql = mysql_connect("localhost","root","root");
if ( !$mysql ) {
foo('query failed', mysql_error());
}
$rc = mysql_select_db("bib", $mysql);
if ( !$rc) {
foo('select db', mysql_error($mysql));
}
$id = "12";
$titlu = "Joe";
$query = "INSERT INTO carte SET id='$id', titlu='$titlu'";
$result = mysql_query($query, $mysql);
// Display an appropriate message
if ($result) {
echo "<p>Product successfully inserted!</p>";
}
else {
foo("There was a problem inserting the Book!", mysql_error($mysql), false);
}
mysql_close($mysql);
function foo($description, $detail, $die=false) {
echo '<pre>', htmlspecialchars($description), "</pre>\n";
if ( defined('DEBUGOUTPUT') && DEBUGOUTPUT ) {
echo '<pre>', htmlspecialchars($detail), "</pre>\n";
}
if ( $die ) {
die;
}
}
try this to connect to database:
$mysqlID = mysql_connect(DB_HOST, DB_USERNAME, DB_PASSWORD) or die("Unable to connect to database");
mysql_select_db(DB_DATABASE) or die("Unable to select database ".DB_DATABASE);
also, try this as your insert query:
$query = "INSERT INTO carte (id, title) values ('".$id."', '".addslashes($titlu)."')
$result = mysql_query($query) or die(mysql_error());
By using die(), it will tell you where it has failed and why