<?php
$link = mysql_connect('localhost', 'sc2broad_testing', '1BananA2');
if (!$link) {
die('Could not connect: ' . mysql_error());
}
echo 'Connected successfully';
mysql_query("INSERT INTO Persons (re) VALUES ('Peter')");
mysql_close($link);
?>
This code isnt taking the value 'peter' and inserting it into persons row 're'?? should i attempt to tell it which database somewhere? thanks . it is saying it connects successfully even if i am not telling it which database to connect to? only the server and user? i am confused.
I think you may need to specify the database that you are querying to?
mysql_select_db('db_name', $link)
If not try changing the mysql_query to:
print("INSERT INTO Persons (re) VALUES ('Peter')");
You can then check the query is correct and test it works outside of the php.
Related
I run this code, and it does not add data to the table. (Also I get no errors).
I think perhaps in someway it can't be sending the data to the table, is there a way to check whats happening in that regard or what do you guys think I should do to narrow down the problem?
I am using netbeans 8 if thats relevant.
<?php
$servername="localhost";
$username="test";
$password="";
$mysqli = new mysqli($servername, $username, $password);
if (!$mysqli){
die('Could not connect: ' . mysql_error());
}
echo 'Connected successfully';
$sql = ("INSERT INTO test_table (column_test) VALUES ('boo')");
?>
You need to execute query, put code below under your $sql
mysqli_query($mysqli, $sql);
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 completly new to databases and am trying to make a simple query to a postgreSQL database. Now the problem is that I don't really understand how things work. In my index.php I do
$dbconn = pg_connect("host=myhost port=5432 dbname=mydbname user=myuser password=mypass sslmode=require options='--client_encoding=UTF8'") or die('Could not connect: ' . pg_last_error());
And I am connected to the database. When I try to query it from another myphp.php
$query = "INSERT INTO public.account_recover_users VALUES ($mykey,$email,$name)";
$result = pg_query($query);
I get the error in the title description. What is the proper way to connect to the database so that I can access it from other files than index.php as well?
P.S. $mykey,$email,$name are 3 variables containing a key (randomly-generated), an email and a name.
I need to retrieve the auto increment field from my database table. I tried the following but $id is always just empty.
The insert works too.
My table is as follows:
idint(9) NOT NULL auto_increment,
and id is set as primary
What am I doing wrong?
$conn = mysql_connect($host,$username,$password);
mysql_select_db($database, $conn) or die( "Unable to select database");
include "update_activity.php";
updateActivity("logged in", "On Break");
$date = date("m/d/y"); $starttime = time();
$sesh = $_SESSION['fname']." ".$_SESSION['lname'];
$q = "INSERT INTO `breaks` (date, starttime, user) VALUES ('".$date."', '".$starttime."', '".$sesh."')";
$query = mysql_query($q, $conn);
$id = mysql_insert_id($conn);
echo var_dump($id); exit;
edited to show my more recent attempts
Have read all comments given and your replies to each.
Only one of these is possible:
Either the query works properly OR
You are not getting the generated primary key.
Both of these can never be true.
Define, how you know query is working? Do you know the max PK before and after the running query? Is the insert happening from some other place or thread or even other user? the query is working properly from code or from your mysql client?
To diagnose the problem, we have to go though the normal way.
Dump your generated query before calling mysql_query.
Wrap a error checking system around your query call so php can tell you if the query worked or not. I am sure just by these two steps you will realize the root cause of the problem.
error_reporting(E_ALL);
ini_set('display_errors','on');
echo "before calling: $q\n";
$query = mysql_query($q, $conn);
if(!$query)
{
echo "Error:" . mysql_error($conn);
return;
}
echo " generated id:" . mysql_insert_id($conn);
#adelphia as far as i get the idea there is a problem in the query that is executed.
plz check the query properly
Borrow a lead from this code extracted from here:
http://php.net/manual/en/function.mysql-insert-id.php
<?php
$link = mysql_connect('localhost', 'mysql_user', 'mysql_password');
if (!$link) {
die('Could not connect: ' . mysql_error());
}
mysql_select_db('mydb');
mysql_query("INSERT INTO mytable (product) values ('kossu')");
printf("Last inserted record has id %d\n", mysql_insert_id());
?>
The problem with your insert query
$q = "INSERT INTO `breaks` (date, starttime, user)
VALUES ('".$date."',
'".$starttime."',
'".$_SESSION['fname'] $_SESSION['lname']."')";
try with this
and main thing you are using most of the deprecated "mysql" things like "mysql_insert_id()"
store the values that u want to pass into an array or variable and pass it in the insert query.
its should work fine then...
I have been able to manually insert values in my table using phpmyadmin, and even if i end up using the same php code i get from php my admin to call the query it STILL won't add the value to the table. here is the code:
<?php
$link = mysql_connect('localhost', 'username', 'password');
if (!$link) {
die('Could not connect: ' . mysql_error());
}
echo 'Connected successfully';
mysql_select_db('sc2broating1', $link);
$sql = "INSERT INTO `sc2broad_tesing1`.`Persons` (`re`) VALUES (\'hello11\')";
mysql_query($sql);
mysql_close($link);
?>
Don't escape value.
$sql = "INSERT INTO `sc2broad_tesing1`.`Persons` (`re`) VALUES ('hello11')";
I would also consider using bound parameters, as seen in mysqli::prepare, if Mysqli is an option.