I have a MSSQL database. I wrote a php page to connect and retreive data from it. The code works on my localhost.
$strCon = "Provider=SQLOLEDB;Data Source=local;database=test;uid=sa;pwd=go;";
$strAlias = "GEORGE";
$strSql = "SELECT LINK FROM LINKS WHERE ALIAS = '" . $strAlias . "'";
$Con = new COM ("ADODB.Connection") or die("Cannot start ADO");
$Rs = new COM ("ADODB.Recordset") or die("Cannot start ADO");
$Con->open($strCon);
$Rs->open($strSql,$Con,1,3);
if (!$Rs->EOF && !$Rs->BOF) {
$strTargetLink = $Rs->Fields['LINK'];
echo $strTargetLink;
}
$Rs->Close();
$Con->Close();
$Rs = null;
$Con = null;
When I run this code on the server, I receive some error? when echo $strTargetLink; row is executed, word
Object
is received on the sent html page and also in the source code of that page.
I am running PHP on IIS as a FastCGI application. Both PHP 4.4.7 and 5.2.6 are supported.
Any ideas? What does this Object text mean?
Thanks.
Instead of using echo, I now tried
var_dump($strTargetLink);
and received the information
object(COM)(1) { [0]=> resource(3) of type (COM) }
try using var_dump instead of echo. it will at least give you more information about the object contained in $strTargetLink.
I have found a solution and now it works. So just in case you would like to connect with ADO, I am writing what I have done.
Instead of
$strTargetLink = $Rs->Fields['LINK'];
I wrote
$strTargetLink = $Rs->Fields(0);
Related
I've successfully connected to a remote IBM i DB2 database (AS400) from my local Windows PC via PHP. I'm using the IBM Data Server Client in conjunction with the db2_* functions in PHP. The problem I'm having is that despite my library list being set properly, it is not being used for unqualified table names. Instead it uses the current user name as the library. However, when I qualify the table names everything works like a charm.
I've confirmed that my library list is actually changing when I create the connection by querying QSYS2.LIBRARY_LIST_INFO.
$database = '<database name>';
$user = '<user name>';
$password = '<password';
$port = <port>;
$options['i5_naming'] = DB2_I5_NAMING_ON;
$options['autocommit'] = DB2_AUTOCOMMIT_OFF;
$options['i5_libl'] = 'MYLIB YOURLIB ANYLIB';
$conn = db2_connect($database, $user, $password, $options);
if ($conn) {
echo "Connection succeeded."; //It succeeds
}
else {
echo db2_conn_error()." | ".db2_conn_errormsg()."<br />";
echo "Connection failed.";
}
$sql = "SELECT * FROM QSYS2.LIBRARY_LIST_INFO";
//Works and proves my library list reflects
//what I passed in when creating the connection.
//$sql = "SELECT * FROM LIBRARY_LIST_INFO";
//Generates: "42S02 : [IBM][CLI Driver][AS] SQL0204N "<user name>.LIBRARY_LIST_INFO" is an undefined name. SQLSTATE=42704 SQLCODE=-204"
//where <user name> is the username used to connect to the DB.
//It should be using the library list specified when creating the connection though.
//This holds true for any table from any library including those specified
//when creating the connection (which includes QSYS2).
$stmt = db2_prepare($conn, $sql);
$result = db2_execute($stmt);
if($result){
while($row = db2_fetch_assoc($stmt)){
echo "<pre>";
var_dump($row); //In addition to entries for QSYS, QSYS2, QUSRSYS and QHLPSYS I get entries for MYLIB, YOURLIB and ANYLIB.
echo "</pre>";
}
}else{
echo "failed<br />";
echo db2_stmt_error()." : ".db2_stmt_errormsg()."<br />";
}
Has anyone ever run into this while enabling i5_naming when connecting to a remote DB2 server? I'm not really sure why it wouldn't be using my library list as the PHP manual states "Unqualified files are resolved using the library list for the job." when enabled. http://php.net/manual/en/function.db2-connect.php
I finally solved this after opening a PMR with IBM. All I had to do was apply the latest Fix Pack for DB2 Connect Personal Edition.
Suggested Fix Packs for DB2 Connect:
http://www-01.ibm.com/support/docview.wss?rs=71&uid=swg21321001
Basically the DB2 Connect version I had was released prior to 2013. It was in 2013 IBM added two tier support by adding the i5_naming option. So my DB2 Connect setup was effectively ignoring the option I was passing. So that explains why the other options still went through. On the DB side, since it didn't receive a value for i5_naming - it remained as the default.
Hey I am iOS developer I am trying to create simple JSON output from my website. I found good start link and here is some explanation how to do it.
So I've created accounts.php file and put it to my public_html folder
<?php
include_once("JSON.php");
$json = new Services_JSON();
$link = mysql_pconnect("localhost", "user", "pass") or die("Could not connect");
mysql_select_db("iglobe") or die("Could not select database");
$arr = array();
$rs = mysql_query("SELECT * FROM users");
while($obj = mysql_fetch_object($rs)) {
$arr[] = $obj;
}
Echo $json->encode($arr);
?>
Of course I use my user and password and I pointed my just created database ob my end.
so when I try to request my file so http//mywebsite.com/accounts.php there is no data.
I tried to use google chrome and Postman so it says No response received when I switch to JSON. For HTML there is no info in Postman.
My question how can I test it? even if I use Echo(123) before include_once("JSON.php"); line there is no 123 on html page.
I tried to test PHP with only this code:
<?php
phpinfo();
?>
and it works. I have PHP Version 5.4.32
First of all, simply use PHP's function json_encode($arr). It does exactly what you are asking for and is pretty much included in every version of PHP that I can think of.
Documentation
Also, I am not sure if this is the issue, but you may want to change Echo ==> echo. This is generally convention at the very least.
SUPER IMPORTANT
Finally, DO NOT USE mysql extension. Its is dangerous, may not work correctly, and has security vulnerabilities. Use mysqli or PDO.
Matrosov -
You are very close. Use the json_encode function to output your code via the PHP manual. Also consider using mysqli instead of mysql for your database connection as it has been better support for modern MySQL servers.
http://php.net/manual/en/function.json-encode.php
http://php.net/manual/en/book.mysqli.php
<?php
include_once("JSON.php");
$link = mysqli_connect("localhost", "user", "pass") or die("Could not connect");
$link->mysql_select_db("iglobe") or die("Could not select database");
$arr = array();
$rs = mysql_query("SELECT * FROM users");
while($obj = mysql_fetch_object($rs)) {
$arr[] = $obj;
}
echo json_encode($arr);
?>
Hi I know this is a little general but its something I cant seem to work out by reading online.
Im trying to connnect to a database using php / mysqli using a wamp server and a database which is local host on php admin.
No matter what I try i keep getting the error Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, boolean given when i try to output the contents of the database.
the code im using is:
if (isset($_POST["submit"]))
{
$con = mysqli_connect("localhost");
if ($con == true)
{
echo "Database connection established";
}
else
{
die("Unable to connect to database");
}
$result = mysqli_query($con,"SELECT *");
while($row = mysqli_fetch_array($result))
{
echo $row['login'];
}
}
I will be good if you have a look at the standard mysqli_connect here
I will dont seem to see where you have selected any data base before attempting to dump it contents.
<?php
//set up basic connection :
$con = mysqli_connect("host","user","passw","db") or die("Error " . mysqli_error($con));
?>
Following this basic standard will also help you know where prob is.
you have to select from table . or mysqli dont know what table are you selecting from.
change this
$result = mysqli_query($con,"SELECT *");
to
$result = mysqli_query($con,"SELECT * FROM table_name ");
table_name is the name of your table
and your connection is tottally wrong.
use this
$con = mysqli_connect("hostname","username","password","database_name");
you have to learn here how to connect and use mysqli
Can anyone see what the problem with my code is / where im going wrong?
I know i have the correct host,database,user and password.
This is the code in the php file, it should get all the details available on the players from my sql database, however if i go on the page it just gives me a white page. Im using go daddy as a host and my database is also on there.
Any ideas? thanks
<?php
$host = "abc12345"; //Your database host server
$db = "abc12345"; //Your database name
$user = "abc12345"; //Your database user
$pass = "abc12345"; //Your password
$connection = mysql_connect($host, $user, $pass);
//Check to see if we can connect to the server
if (!$connection) {
die("Database server connection failed.");
} else {
//Attempt to select the database
$dbconnect = mysql_select_db($db, $connection);
//Check to see if we could select the database
if (!$dbconnect) {
die("Unable to connect to the specified database!");
} else {
$query = "SELECT * FROM Player";
$resultset = mysql_query($query);
$records = array();
//Loop through all our records and add them to our array
while ($r = mysql_fetch_assoc($resultset)) {
$records[] = $r;
}
//Output the data as JSON
echo json_encode($records);
}
}
?>
The script is all right, I checked it with a different query.
Assuming that the table Player is not empty, the error can be either with your raw sql query (as pointed by #Sharikov in comments), otherwise the error is with the way you have configured your godaddy server.
For debugging that, I would suggest inserting dummy print_r or echo statements before you execute the query, and going through your apache logs at /var/log/apache2/access.log.
Also make sure that you don't have any core php package missing on your server (like php5-mysql if you use mysql).
I want to retrieve or output data in the database but I kept on getting the error called "Resource ID".
Here is my code:
<?php
$host="localhost";
$username="root";
$password ="123192";
$db_name = "customers";
//Connecting to your Host
mysql_connect("$host","$username","$password") or die("Failed To Connect The server");
//Selecting your Database
mysql_select_db("$db_name") or die("Failed To Select The DB");
$name = $_REQUEST['customerName'];
echo 'WELCOME! <b>'.$name.'</b> We hope that you\'ll Enjoy your stay ';
$sql="SELECT Name FROM `people` WHERE id =2 && Name = 'Kyel'";
$rs=mysql_query($sql);
echo "$rs";
?>
If I need improvement regarding my code kindly tell me.
mysql_query() returns a resource. The to string (implicitly triggered by using echo to output it) of that is Resource ID # followed by the id.
A resource in PHP is only supposed to be used with other PHP functions. This includes but is not limited to file, curl, ftp handles, etc.
I could tell you to..
(a) use mysql_fetch_array() (or similar) or
(b) use PDO.
The latter is by far much better advice.
Try this instead of the echo statement:
$array = mysql_fetch_assoc($rs);
var_dump ($array);