I am using wamp server 2.0 working with PHP and my database is Oracle 10g.
I am new to php and I am try to fetch data from database.
There are two columns in my table. I want show 1 column data.
After executing I get only blank page, without data from database (and yes, there is data in my database).
How can I fix this?
<?php
$c = oci_connect("system", "123", "localhost/XE");
if (!$c) {
echo "Unable to connect: " . var_dump( oci_error() );
die();
}
$s = oci_parse($c, "select col2 from tab1");
oci_execute($s, OCI_DEFAULT);
while ($row =oci_fetch($s)) {
echo $row['name']."<br>";
}
// Commit to save changes...
oci_commit($c);
// Logoff from Oracle...
oci_free_statement($s);
oci_close($c);
?>
oci_fetch copies a result into an internal buffer that you have to access e.g. using oci_result.
BTW: Had you set a higher value for error_reporting and turned on display_errors, then you would have noticed, that you tried to treat a boolean as an array when printing $row['name'].
Related
I have a web page created in php using html code. I want to save user information entered in my web page to a MySQL database. I am using php as the middle man to link the frontend web page(htmnl code) to the database(mysql).
Inside my link folder (middle man php file) I have the following:
<?php
//Gets server connection credentials stored in serConCred2.php
require_once('ConCred2.php');
//SQL code for connection w/ error control
$con = mysqli_connect(DB_HOST, DB_USER, DB_PASSWORD, DB_NAME);
if(!$con){
die('Could not connect: ' . mysqli_connect_error());
}
//Selection of the databse w/ error control
$db_selected = mysqli_select_db($con, DB_NAME);
if(!$db_selected){
die('Can not use ' . DB_NAME . ': ' . mysqli_error($con));
}
//Co-PI and Co-Investigator Information variables
$Co_FNAME = $_POST['fname'];
$Co_LNAME = $_POST['lname'];
$Co_SLNAME = $_POST['slname'];
$Co_DEGREE = $_POST['Degree_Selection'];
$Co_DEGREE_Other = $_POST['other_specify_degree']; //hold the value of degree if user selected other from the dropdown menu
$Co_CPOS = $_POST['Current_Position_Selection'];
$Co_CPOS_Other = $_POST['other_specify_cpos']; //hold the value of Current Position if user selected other from the dropdown menu
$Co_INST = $_POST['Institution_Selection'];
$Co_INST_Other = $_POST['other_specify_inst']; //hold the value of Current Position if user selected other from the dropdown menu
$Co_SCHOOL = $_POST['School_Selection'];
$Co_SCHOOL_Other = $_POST['other_specify_school']; //hold the value of Current Position if user selected other from the dropdown menu
$Co_DEPART = $_POST['Department_Selection']; //Este se estara eliminando en la version online
$Co_DEPART_Other = $_POST['other_specify_department']; //hold the value of Department if user selected other from the dropdown menu
$Co_PROGRAM = $_POST['program'];
$Co_EMAIL = $_POST['email'];
$Co_PHONE = $_POST['phone'];
//If decition when user select other from the dropdown menu
if($Co_DEGREE == "other_degree") $Co_DEGREE = $Co_DEGREE_Other;
if($Co_CPOS == "other_cpos") $Co_CPOS = $Co_CPOS_Other;
if($Co_INST == "other_inst") $Co_INST = $Co_INST_Other;
if($Co_SCHOOL == "other_school") $Co_SCHOOL = $Co_SCHOOL_Other;
if($Co_DEPART_Other == "other_department") $Co_DEPART = $Co_DEPART_Other;
//This sets a starting point in the rollback process in case of errors along the code
$success = true; //Flag to determine success of transaction
//start transaction
echo "<br>1. Going to set autocommit to 0";
$command = "SET AUTOCOMMIT = 0";
echo "<br>2. Autocomint has been set to 0";
echo "<br>3. Going to run query to see if result is true or false";
$result = mysqli_query($con, $command);
echo "<br>4. Finished running the query. Result is:" . $result;
echo "<br>5. Going to set command to BEGIN";
$command = "BEGIN";
echo "<br>6. Command is now BEGIN";
echo "<br>7. Going to run query for command BEGIN";
$result = mysqli_query($con, $command);
echo "<br>8. Query runned for command BEGIN";
echo "<br>9. Result value is: " . $result;
//Saves Pi values into database
/**
$sqlCoPI = "INSERT INTO co_pi_table (Fname, Lname, SLname, Degree, Current_Position, Institution, School, Department, Program, Email, Phone)
VALUES('$Co_FNAME', '$Co_LNAME', '$Co_SLNAME', '$Co_DEGREE', '$Co_CPOS', '$Co_INST', '$Co_SCHOOL', '$Co_DEPART', '$Co_PROGRAM', '$Co_EMAIL', '$Co_PHONE')";
*/
echo "<br>10. Going to write sql command to populate table pi_table";
/**
$sqlPi = "INSERT INTO pi_table (Fname, Lname, SLname, Degree, Current_Position, Institution, School, Department, Program, Email, Phone)
VALUES('$Co_FNAME', '$Co_LNAME', '$Co_SLNAME', '$Co_DEGREE', '$Co_CPOS', '$Co_INST', '$Co_SCHOOL', '$Co_DEPART', '$Co_PROGRAM', '$Co_EMAIL', '$Co_PHONE')";
*/
$sqlPi = "INSERT INTO pi_table (Fname) VALUES('$Co_FNAME')";
//Checks to see if theres an error in the pi db con
echo "<br>11. Sql command finished writting.";
echo "<br>12. Going to query the sql finished command to the database to determine value of result.";
$result = mysqli_query($con, $sqlPi);
echo "<br>13. Finished running sql command to database. Result value is: " . $result;
echo "<br>14. Going to enter if statements depending on result value";
if($result == false){
//die ('<br>Error in query to PI table: ' . mysqli_error($con));
echo "<br>15. I am inside the false statement. Success is going to be set as false. ";
$success = false;
//$success = true; //Cahnged this in order to test if values are being saved to db. Change back to false.
}
//Checks for errors or craches inside the code
// If found, execute rollback
echo "<br>16. Going to verify is success is true.";
if($success){
$command = "COMMIT";
$result = mysqli_query($con, $command);
//echo "<br>Tables have been saved with 0 errors.";
echo "<br><p style=\"color: red;\"Principal Investigator has been saved successfuly. <br><br>
You may now CLOSE this page and press the<br><br> \"Refresh List\" <br><br>
button to display name in dropdown menu selection.</p>";
}
else{
$command = "ROLLBACK";
$result = mysqli_query($con, $command);
echo "<br>17. Success was determined to be false.";
echo "<br>Error! Databases could not be saved.<br>
Contact system manager to report error. <br> <br>" . mysqli_error($con);
}
echo "<br>18. Setting autocommit back to 1 again.";
$command = "SET AUTOCOMMIT = 1"; //return to autocommit
$result = mysqli_query($con, $command);
//Displays message
//echo '<br>Connection Successfully. ';
//echo '<br>Database have been saved';
//Close the sql connection to dababase
mysqli_close($con)
?>
As you can read, I am requiring users to fill out their information. Some of the information required are dropdown menu fields that user selects an option from among the presented ones.
The problem I am having is, when the above php code executes, it determines that the $result variable is false and doesn't save anything. When you execute the code, you get the following messages displayed:
1. Going to set autocommit to 0
2. Autocomint has been set to 0
3. Going to run query to see if result is true or false
4. Finished running the query. Result is:1
5. Going to set command to BEGIN
6. Command is now BEGIN
7. Going to run query for command BEGIN
8. Query runned for command BEGIN
9. Result value is: 1
10. Going to write sql command to populate table pi_table
11. Sql command finished writting.
12. Going to query the sql finished command to the database to determine value of result.
13. Finished running sql command to database. Result value is:
14. Going to enter if statements depending on result value
15. I am inside the false statement. Success is going to be set as false.
16. Going to verify is success is true.
17. Success was determined to be false.
Error! Databases could not be saved.
Contact system manager to report error.
18. Setting autocommit back to 1 again.
For security purposes I cant post the html content since it has sensitive name information nor the databases. Although I can ensure that the tables inside the database are called exactly as mentioned in the sql command line.
I HAVE FOUND THE PROBLEM!
After long debating I decided to recreate the database In which all the information was being stored. When I redirected the table in my sql command ( Instead of saving it in "pi_table" I saved it in a newly created database called "pi_table_2") and everything worked out properly.
Aparently my database got corrupted and phpMyAdmin didn't recognized that it was curropted.
For reference my database tables where in InnoDB format. What might have cause this to happen, who knows but if you ever encounter a similar problem, creating a small testing database and see if it saves. If it does, recreate the table and it might solve your issue like it solved mine.
Once again thank you a lot guys!!!!!
I am looking at the code and everything seems to be in order, could be a syntax error like a missing quotation for example:
//SQL code for connection w/ error control
$con = mysqli_connect("DB_HOST", "DB_USER", "DB_PASSWORD", "DB_NAME");
also
$db_selected = mysqli_select_db($con, "DB_NAME");
or die ("Cant select Database");
}
Hope this help.
Cheers;
Hasan
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´m pretty new to php and i have encountered a problem with no idea where it comes from.
I want to connect to an sql-db via php in order to access this via iOS. But the first step fails allready :(
This is a screenshot with login data from my provider:
Then i uploaded this php to the website:
<?php
// Create connection
$con=mysqli_connect("db559233526.db.1and1.com","dbo559233526","Correct Password","db559233526");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
// This SQL statement selects ALL from the table ‚Cards‘
$sql = "SELECT * FROM Cards";
// Check if there are results
if ($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($con);
?>
You con see the result here:
http://sektor3d.de/service.php
Now i wonder, why the last half of the code is shown as text? And why is the "break" at that point?
Thanks for help or any idea!
Take a look at your page source, your server seems to not be able to run PHP. If you're convinced it can, add this code to your .htaccess file:
AddType application/x-httpd-php .php
Check if your server support PHP
Create a file (in any text editor) and inside the file type:
<?php
phpinfo();
?>
Save it as info.php
Upload it to your server, and then, in your web browser, navigate to that file eg. http://sektor3d.de/info.php
We have a website that accesses a SQL Server 2005 server for one query. Currently, the site is in ASP, we are moving it to PHP, and the PHP one is currently being tested. After we run a few successful queries on the PHP site, it returns the "Error in database query. Please try again later" line in the code below. When I rewrote that line with sqlsrv_errors to elaborate, it told me that the table didn't exist. There are about 40 tables in the database, but after the error happens it only shows 8 of them in Management Studio. However, if I allow it to sit for about 5 minutes, all of the tables are restored. No matter how many times the old ASP site is used, the table does not do this. However, when the tables disappear from using the new site, the old site shows inaccessible for a few minutes until the tables re-appear in SQL Server management studio. I didn't see any kind of connection limits on the SQL Server, so I don't know whether it's something I'm doing in the PHP SQL queries or within the SQL Server properties.
<?php
include ("dbvals.inc.php");
if (!empty($_POST['lastnamebox'])) {
$dbhandle = sqlsrv_connect($dbServer, $connectioninfo);
if($dbhandle == false){
echo "Error connecting to database. Please try again later. ";
}
else{
$query = "SELECT * FROM Person WHERE LastName LIKE '%' + ? +
'%' AND InactiveFlag = 'N' ORDER BY LastName, FirstName";
$params = array();
array_push ($params, $_POST['lastnamebox']);
$results = sqlsrv_query($dbhandle, $query, $params);
if($results == false){
echo "Error in database query. Please try again later.";
//This is printed when database tables temporarily disappear
}
else
{
$row = sqlsrv_fetch_array( $results, SQLSRV_FETCH_ASSOC);
if($row){
do{
echo "<tr><td class='tablecell'>";
echo $row['LastName'] . "," . $row['FirstName'] . "<br>";
echo "Address: " . $row['Address'] . "<br>";
echo "City, State, Zip: " . $row['CSZ'] . "<br>";
echo "</tr></td'>";
}while($row = sqlsrv_fetch_array( $results, SQLSRV_FETCH_ASSOC));
}
else{
echo "No results found. Please try another query.";
}
}
}
sqlsrv_free_stmt($results);
sqlsrv_close($dbhandle);
}
else {
echo "Please type a value in the search box.";
}
?>
The $query is wrong to begin with. Change it to:
$query = "SELECT * FROM Person WHERE LastName LIKE '%' + ? +
'%' AND InactiveFlag = 'N' ORDER BY LastName, FirstName";
Not sure about the tables disappearing though. What driver are you using?
Just another issue I'm seeing with the code not sure if it's related to your problem. or another copy error but here it is.
if ($r1 = sqlsrv_fetch_array($results)) {
while( $row = sqlsrv_fetch_array( $results, SQLSRV_FETCH_ASSOC)){
The first fetch should also have '($results,, SQLSRV_FETCH_ASSOC)'
Additional that check is going to eat the first returned record which may or may NOT be what you intended.
Also it is possibly that PHP seeing some results in the while as false even though they aren't and the server is still waiting on you to finish getting the rest. Seen code like that cause 'Server has got away' errors in MySQL which could be what is going on here as well.
Not sure of the syntax but a sqlsrv_clode_cursor() just before the connection close might also fix your issue if there's some kind of connection polling going on. Could be simply running out of connection or getting old one in a incorrect state.
SELECT * FROM Person WHERE AND LastName LIKE
Looks like there something missing between the WHERE and AND to me.
I don't know SQL Server really so it could allow that but it's not standard SQL. Depending on how it's reacting to that error I could see it 'Going away' so to say and reporting tables missing.
I'm trying to use jQuery-jTable to list data from a MS-Access database through PHP.
I want to change the sample provided by jTable.org : http://www.jtable.org/downloads/jTable-PHP-Samples.zip
<?php
try
{
//Open database connection
$db_connection = odbc_connect("Persist Security Info=False;DRIVER={Microsoft Access Driver (*.mdb)}; DBQ=c:\websiagh\books\54.mdb", "ADODB.Connection", "password", SQL_CUR_USE_ODBC) or die('Cannot connect to 54');
//Getting records (listAction)
if($_GET["action"] == "list")
{
//Get records from database
$query = 'SELECT * FROM asnad WHERE (sanadno Between 10 AND 20 )';
$result = odbc_exec($db_connection , $query );
//Add all records to an array
$rows = array();
while( $row = odbc_fetch_array( $result ) )
{
$rows[] = $row;
}
//Return result to jTable
$jTableResult = array();
$jTableResult['Result'] = "OK";
$jTableResult['Records'] = $rows;
print json_encode($jTableResult);
}
...
The rest of the code is not changed.
The code ( query execution and fetching data using odbc ) works well when not using jTable. However when using it in the code as above , I get this error :
An error occured while communicating to the server.
But when I export the data from MS-Access to mysql, there is no problem at all.
For me your query execution and fetching data using odbc didn't work. This may be depending on the Window OS and the version of PHP.
The following however did work with jTables (credits to w3schools.com):
Open the Administrative Tools icon in your Control Panel.
Double-click on the Data Sources (ODBC) icon inside.
Choose the System DSN tab.
Click on Add in the System DSN tab.
Select the Microsoft Access Driver. Click Finish.
In the next screen, click Select to locate the database. Give the
database a Data Source Name (DSN).
Click OK.
After, to create a connection use the following (assuming your database is called 54.mdb):
$db_connection=odbc_connect('54','','');
if (!$db_connection) {
exit("Connection Failed: " . $db_connection);
}
The rest of the code remains as is.
Hope this helps.