I used
mysqli_connect("infos in here");
at the top of my page, and tried to use
mysqli_query("INSERT INTO and other info here");
When I do that, I get this error:
Warning: mysqli_query() expects at least 2 parameters, 1 given in (...)
But if I instead use
$con = mysqli_connect("infos in here");,
$mysqli_query($con,"INSERT INTO and other info here");
The error goes away, and my script works.
My problem is that I need to use mysqli_query two different times in my page, and I don't want to open the connection again when it's already open.
How can I handle this?
Thanks.
My problem is that I need to use mysqli_query two different times in
my page, and I don't want to open the connection again when it's
already open.
How is it a problem ? open once query as many times then close the connection, example:
<?php
$con=mysqli_connect("localhost","my_user","my_password","my_db");
// Check connection
if (mysqli_connect_errno()){
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
// Perform queries
mysqli_query($con,"SELECT * FROM Persons");
//one more
mysqli_query($con,"INSERT INTO Persons (FirstName,LastName,Age)
VALUES ('Glenn','Quagmire',33)");
mysqli_close($con);
?>
No need to connect two times. An example here..
$con = mysqli_connect("host", "user", "password", "db");
if(mysqli_connect_errno()){
die(mysqli_connect_error());
}
//Query one
$result1 = mysqli_query($con, 'Query String');
//Query Two
$result2 = mysqli_query($con, 'Query String'); //Used same $con variable
//After finishing all queries
mysqli_close($con);
If you need to select multiple database then
$con = mysqli_connect("host", "user", "password");
//for DB bd_name1
mysqli_select_db($con, 'bd_name1');
$result1 = mysqli_query($con, 'Query String');
//for DB bd_name2
mysqli_select_db($con, 'bd_name2');
$result2 = mysqli_query($con, 'Query String');
But not connect frequent time.
Related
i'm in a middle of making a query with my new database and i keep getting the message (Database query failed) through the code below:
<?php
//create a database connection
$dbhost= "localhost";
$dbname= "widget_corp";
$connection=mysqli_connect($dbhost , $dbname);
if(mysqli_connect_errno()){
die("Database connection failed :" . mysqli_connect_error ."(". mysqli_connect_errno .")");
}
?>
<?php
//perform a database query
$query = "SELECT * FROM subjects";
$result = mysqli_query($connection ,$query);
if (!$result){
die("Database query failed.");
}
?>
please advise
You aren't passing in a username or password. mysqli_connect() requires four parameters to be passed in : http://php.net/manual/en/function.mysqli-connect.php
You're only passing in a host and database name.
This question already has answers here:
Can I mix MySQL APIs in PHP?
(4 answers)
Closed 6 years ago.
Can you figure out my code? All the code does is: No database selected It won't get the data from the db. The server os is Ubuntu or OS X. I been pulling my hair out for hours.
<?php
mysqli_connect("localhost", "root", "");
mysql_select_db("hit-counter");
$sql_get_count = mysql_query("SELECT id FROM hit_info ORDER BY id DESC LIMIT 1");
if($sql_get_count === FALSE) {
die(mysql_error());
}
while($row = mysql_fetch_assoc($sql_get_count)) {
print_r($row);
}
?>
I try this, it does the same
<?php
mysql_connect("localhost", "root", "");
mysql_select_db("hit-counter");
$sql_get_count = mysql_query("SELECT id FROM hit_info ORDER BY id DESC LIMIT 1");
if($sql_get_count === FALSE) {
die(mysql_error());
}
while($row = mysql_fetch_assoc($sql_get_count)) {
print_r($row);
}
?>
you have an error in your code. You use mysqli_ function to connect the server but you use a deprecated function mysql_ to select the database.
Try this code:
mysqli_connect("localhost", "root", "");
mysqli_select_db("hit-counter");
Another option when using mysqli_ is to select the database you want during connecting to the server:
$link = mysqli_connect("127.0.0.1", "my_user", "my_password", "my_db");
You didn't mention database name:try this
<?php
$con = mysqli_connect("127.0.0.1","root","654321","testV2") or die("Some error occurred during connection " . mysqli_error($con));
// Write query
$strSQL = "SELECT id FROM did ORDER BY id DESC LIMIT 1";
// Execute the query.
$query = mysqli_query($con, $strSQL);
while($result = mysqli_fetch_array($query))
{
echo $result["id"]."
";
}
// Close the connection
mysqli_close($con);
?>
You cannot interchange the mysql and mysqli functions, please modify your mysql_select_db to mysqli_select_db.
I will not go over the errors everyone else has pointed out. But, I will mention one that no one has. I think the - character in your database name will also cause problems. You should enclose the database name in back ticks. The back tick is this ` character, most likely the far left key above the TAB key. If you had error reporting turned on, or looked at your php error log, you would have seen the error.
I´m trying to update some different registers in a mysql database sending the commands from a FOR loop in php, but the query is only done the 1st loop. Here´s the code:
$conexion = mysql_connect($hostname, $user, $pass) or die ("Error establishing connection with the Database");
mysql_select_db($db,$conexion) or die("Error selecting the Database");
$j=0;
for ($i=0;$i<count($notifs);$i++){
$sql="UPDATE tef SET notif='$notifs[$i]' WHERE sn_rec='$unsersn_recs[$j]';";
echo $sql."<br>";
$res=mysql_query($sql, $conexion) or die (mysql_error());
$j++;
}
mysql_close($conexion);
The query text is correctly done (the echo shows the different lines created), but the changes in the database are done only in the 1st loop (1st query) and I don´t receive any error. What may I be missing?
Thanks in advance!
This is wonderful example where you should use prepared statements.
I give you an example which is also secure against SQL injections.
$mysqli = new mysqli($hostname, $user, $pass, $db);
if (mysqli_connect_errno()) {
die("Error establishing connection!");
}
$stmt = $mysqli->prepare("UPDATE tef SET notif=? WHERE sn_rec=?");
$j=0;
for ($i=0;$i<count($notifs);$i++) {
$stmt->bind_param('ii', $notifs[$i], $unsersn_recs[$j]);
$stmt->execute();
if(!empty($stmt->error)) echo $stmt->error;
$j++;
}
$stmt->close();
$mysqli->close();
Hint: If notif or sn_rec are varchar/text types, just replace the 'i' with a 's' in bind_param().
I have a few string variables I am trying to insert them into my DB but I am having trouble because nothing is being inserted into the DB. I know the variables are populated. Since all variables are string I'm converting some of them to integers because those fields in the db table are type integer. I tried assigning the mysql_query to a variable and then check to return an error but it didn't display anything. I'm a bit new at PHP so I'm not sure what's wrong with my code below. I appreciate the help.
$connect = mysql_connect("localhost", "user", "pass");
if (!$connect) { die("Could not connect: ". mysql_error()); }
mysql_select_db("dbname");
mysql_query($connect,"INSERT INTO table1 (id, AU, TI, JO, VL, ISS, PB, SN, UR, DO, SP, EP, PY) VALUES ('NULL', '".$authors."', '".$title."', '".$journal."', '".(int)$volume."', '".(int)$issue."', '".$publisher."', '".$serial."', '".$url."', '".$doi."', '".(int)$startpage."', '".(int)$endpage."', '".(int)$year."')");
mysql_close($connect);
Try to debug your code, adding some more useful checks.
$link = mysql_connect("localhost", "user", "pass");
if (!$link) {
die("Could not connect: ". mysql_error());
}
$dbSelected = mysql_select_db("dbname", $link);
if (!$dbSelected) {
die ("Can't select db: " . mysql_error());
}
$result = mysql_query("YOUR_QUERY", $link);
if (!$result) {
die("Invalid query: " . mysql_error());
}
ps: you may want to use mysqly::query, just because mysql_query is deprecated
ps2: you should google about SQL Injection, since your statement doesn't look secure (unless those values are escaped somewhere)
NOTE: I just noticed that you are using a wrong order for the parameters on mysql_query($query, $link). You have put $link as first parameter.
I am getting the error:
Warning: mysqli_num_rows() expects parameter 1 to be mysqli_result, boolean given in /home/mjcrawle/public_html/home/index.php on line 23
Line 23 turns out to be $num_results = mysqli_num_rows($result); but I am thinking the error is further up but I am having trouble finding it.
The actual code that I am using to connect to the DB is (I do understand there is a redundancy if the database cannot connect):
Any help would be wonderful and a reason for the error would be awesome!
/*Connect To DB*/
$conn = mysqli_connect($host, $user, $pwd)
or die("Could not connect: " . mysql_error()); //connect to server
mysqli_select_db($conn, $database)
or die("Error: Could not connect to the database: " . mysql_error());
/*Check for Connection*/
if(mysqli_connect_errno()){
/*Display Error message if fails*/
echo 'Error, could not connect to the database please try again later.';
exit();
}
/* Query for states */
$query = "SELECT StateAbbreviation, StateName, FROM USState ORDER BY StateName";
$result = mysqli_query($conn, $query);
$num_results = mysqli_num_rows($result);
?>
You have an extra comma before the FROM in query = "SELECT StateAbbreviation, StateName, FROM USState ORDER BY StateName";, you may be getting an error and not having a result when you execute the query.
If the query fails, mysqli_query returns boolean false
After $result = mysqli_query($conn, $query);, you should test the return value before continuing:
if ( ! $result){
$error = mysqli_error($conn);
//do something with the error message
}
See EmCo's answer for why your query is failing.
<?php
$con=mysqli_connect('localhost','root','','dbname') or die ("Connection Failed");
?>
This is the simple method of DB Connection