I am trying to fetch data from a table in oracle database using php 7. I am using OCIFetchInto but when i look to the result array, i find it empty, although it shouldn't be actually empty. knowing that the connection to the database is successful and the table in the database is not empty.
I have debugged the code, and i am getting Resource #4 from the execute function, I have changed the execute function but i still get the same result.
i also changed the data[0] to data['columnname'] but i still cannot retreive it.
Thanks in advance
The code below:
require_once("../include_tse/class_ora_db.php");
$oracle_db = new ora_database("abc");
$error="";
if ($oracle_db->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
echo "Connected successfully";
$Wip=$_REQUEST['rechts'];
echo $Wip;
$sql_query="SELECT Om_Wip_Entity_Name,Quantity,Country_Description FROM abc_print WHERE Om_Wip_Entity_Name= '$Wip'";
$cursor = $oracle_db->execute_sql($sql_query);
$counter=0;
while (OCIFetchInto ($cursor,$row))
{
echo "2222222";
$data[]=$row;
$Wip= $data['0'];
$Quantity= $data['1'];
$Country= $data['2'];
$counter ++;
}
echo "row.$Quantity";
As of PHP 5.4 the ocifetchinto alias has been deprecated (http://php.net/manual/en/function.ocifetchinto.php)
You should change it to something like...
while ($row = oci_fetch_assoc ($cursor))
{
$Wip= $row['Om_Wip_Entity_Name'];
$Quantity= $row['Quantity'];
$Country= $row['Country_Description'];
$data[]=$row;
$counter ++;
}
I'm a bit confused as to how your using the data, but this should give a better idea of how to get it working.
Related
I wrote this php script that allows me to fetch all the rows in a table in my MySQL database.
I have put the echo "1", etc. to see whether it gets to the code at the very end. The output proves it does. However, it does not output anything when echoing json_encode($resultsArray), which I can't seem to figure out why.
Code:
// Create connection
$connection = mysqli_connect("localhost", "xxx", "xxx");
// Check connection
if (!$connection) { die("Connection failed: " . mysqli_connect_error()); } else { echo "0"; }
// select database
if (!mysqli_select_db($connection, "myDB")) { die('Unable to connect to database. '. mysqli_connect_error()); } else { echo "1"; }
$sql = "select * from myTable";
$result = mysqli_query($connection, $sql) or die(mysqli_error($connection));;
echo "3";
$resultsArray = array();
while($row = mysqli_fetch_assoc($result)) {
// convert to array
$resultsArray[] = $row;
}
echo "4";
// return array w/ contents
echo json_encode($resultsArray);
echo "5";
Output:
01345
I figured, it is not about the json_encode, because I can also try to echo sth. like $result['id'] inside the while loop and it just won't do anything.
For testing, I went into the database using Terminal. I can do select * from myTable without any issues.
Any idea?
After around 20hrs of debugging, I figured out the issue.
As I stated in my question, the code used to work a few hours before posting this question and then suddenly stopped working. #MichaelBerkowski confirmed that the code is functional.
I remembered that at some point, I altered my columns to have a default value of an empty string - I declared them as follows: columnName VARCHAR(50) NOT NULL DEFAULT ''.
I now found that replicating the table and leaving out the NOT NULL DEFAULT '' part makes json_encode() work again, so apparently there's an issue with that.
Thanks to everybody for trying anyway!
I need some help with my PHP. I have a trouble with fetching the data from the database. I have hired a PHP developer who did not do his job properly that he have messed up the code which make it don't work so I need some help to fix the issue to get it working again.
When I try this:
//open the database File
$db = new SQLite3('myChannel.db');
if(!$db)
{
echo $db->lastErrorMsg();
}
else
{
$channel_name = $_GET['channels'];
$sql ="SELECT channel, title, start_date, stop_date, description FROM programs WHERE channel='$channel_name'";
$results = $db->query($sql);
while ($row = $results->fetchArray())
{
print_r($row);
}
What happen with the code is it will not fetching the matched data from the database as it will not do anything. I think there is something wrong with the $sql variable.
What I'm expecting to do is I want to look for data in the database where I use the variable called $channel_name, then I want to fetch the matched data to output them in my PHP.
Can you please help me how I can fetch the matched data in the database?
Try this code based on the SQLite PHP docs
class MyDB extends SQLite3 {
function __construct() {
$this->open('myChannel.db');
}
}
$db = new MyDB();
if (!$db) {
echo $db->lastErrorMsg();
} else {
$channel_name = $_GET['channels'];
$sql = "SELECT channel, title, start_date, stop_date, description FROM programs WHERE channel='{$channel_name}'";
$results = $db->query($sql);
while($row = $results->fetchArray(SQLITE3_ASSOC) ) {
print_r($row);
}
}
I changed a few things. I turned your database connection into a class, and I changed your while to include SQLITE3_ASSOC.
Warning: OP's code and as a result this answer has code that is
vulnerable to SQL Injection!
I want to get stings form a mysql server to use as text on my webpage.
That way I can edit the text without editing the html file.
Problem is that the code I have to get the string is quite long, and I don't want to paste it everywhere on the page.
I would also like a tip on how to get just one datafield from the server, and not the whole column like I do here.
So this is what I got. And what I think is to write a function I can call from all the places I want the webpage to get a string or field from the sqlserver. But I don't know how. Can anyone help me?
<?php
$con=mysqli_connect("localhost","user..", "passwd..","db");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$sql = "SELECT topic FROM web_content";
$result = $con->query($sql);
if ($result->num_rows > 0)
{
// output data of each row
while($row = $result->fetch_assoc())
{
echo $row["topic"]. "<br>";
}
} else
{
echo "error";
}
$con->close();
?>
Problem is that the code I have to get the string is quite long, and i
dont want to paste it everywhere on the page.
Put the code into a function, call that function wherever you need to. Then it is just a single line you have to insert.
PHP:
<?php
function connect() {
$con=mysqli_connect("localhost","user..", "passwd..","db");
// Check connection
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
} else {
return $con;
}
}
function renderContent($con) {
$sql = "SELECT topic FROM web_content";
$result = $con->query($sql);
if ($con && ($result->num_rows > 0))
{
// output data of each row
while($row = $result->fetch_assoc())
{
echo $row["topic"]. "<br>";
}
} else {
echo "error";
}
}
HTML:
<?php $con = connect(); ?>
[...]
<div>
<?php renderContent($con); ?>
</div>
[...]
I would also like a tip on how to get just one datafield from the
server, and not the whole coloumn like i do here.
Not the whole column would mean not all rows, but one or some selected ones. That means you are looking for sqls ''WHERE'' clause.
SELECT topic FROM web_content WHERE <where clause>;
Where <where clause> is some clause to narrow down the result set. For example you can narrow down to topics containing some string: ... WHERE topid LIKE '%word%'; or by the IDs are a date range of the entries in your table. You should take a look into the documentation of the query syntax for an explanation: http://dev.mysql.com/doc/refman/5.0/en/select.html
Obviously all of this is just a rough sketch of what you are looking for. Lots of things need improving. Using exceptions for error handling is one thing, just to give an example...
I'm trying to host this php file on my website to connect to a MySQL database.
<?php
// Create connection
$con=mysqli_connect("localhost","username","pass","database");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
// This SQL statement selects ALL from the table 'Locations'
$sql = "SELECT * FROM Locations";
// Check if there are results
if ($result = mysqli_query($con, $sql))
{
// If so, then create a results array and a temporary one
// to hold the data
$resultArray = array();
$tempArray = array();
// Loop through each row in the result set
while($row = $result->fetch_object())
{
// Add each row into our results array
$tempArray = $row;
array_push($resultArray, $tempArray);
}
// Finally, encode the array to JSON and output the results
echo json_encode($resultArray);
}
// Close connections
mysqli_close($con);
?>
And when I try to validate the code using http://writecodeonline.com/php/ it says
Fatal error: Call to undefined function mysqli_connect() on line 2
Searching around said I needed to change the php.ini file but I don't even have one hosted on my site. Is there something wrong with the code?
When I try to access it at www.mydomain.com/service.php it says file not found error... but it's definitely there. I'm working with this tutorial - http://codewithchris.com/iphone-app-connect-to-mysql-database/
You need to install the php-mysql dependency on your server.
yum install php-mysql -y
or your equivalent on your os.
Hi I know this is a little general but its something I cant seem to work out by reading online.
Im trying to connnect to a database using php / mysqli using a wamp server and a database which is local host on php admin.
No matter what I try i keep getting the error Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, boolean given when i try to output the contents of the database.
the code im using is:
if (isset($_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