The code below is working for one if statement but not giving results for another else if and it is showing the 'flights' table in first query but after another condition not display the other table named 'isb to muree'
<?php
$from = isset($_POST['from'])?$_POST['from']:'';
$To = isset($_POST['To'])?$_POST['To']:'';
if( $from =='Islamabad'){
if($To == 'Lahore'){
$db_host = 'localhost';
$db_user = 'root';
$database = 'homedb';
//$table = 'flights';
if (!mysql_connect($db_host, $db_user))
die("Can't connect to database");
if (!mysql_select_db($database))
die("Can't select database");
$result = mysql_query("SELECT * FROM flights");
if (!$result) {
die("Query to show fields from table failed");
}
$fields_num = mysql_num_fields($result);
echo "<h1>Table: 'flights'</h1>";
echo "<table border='1'><tr>";
while($row = mysql_fetch_row($result))
{
echo "<tr>";
// $row is array... foreach( .. ) puts every element
// of $row to $cell variable
foreach($row as $cell)
echo "<td>$cell</td>";
echo "</tr>\n";
}
}
else if( $from =='Islamabad'){
if($To == 'murree'){
if (!mysql_connect($db_host, $db_user))
die("Can't connect to database");
if (!mysql_select_db($database))
die("Can't select database");
$result = mysql_query("SELECT * FROM 'isb to murree'");
if (!$result) {
die("Query to show fields from table failed");
}
$fields_num = mysql_num_fields($result);
echo "<h1>Table: 'isb to murree'</h1>";
echo "<table border='1'><tr>";
while($row = mysql_fetch_row($result))
{
echo "<tr>";
// $row is array... foreach( .. ) puts every element
// of $row to $cell variable
foreach($row as $cell)
echo "<td>$cell</td>";
echo "</tr>\n";
}
}
}
}
mysqli_close($con);
?>
Enclose the table name in backticks (`) instead of single quotes ('):
SELECT * FROM `isb to murree`
To avoid this and other problems in the future, always enclose schema, table, and column names in backticks to distinquish them from data and MySQL keywords. Consider using table names without spaces for easier handling.
As Fred noted, your code is using two database APIs: mysql_* and mysqli_* . In general, the two do not mix well. Consider upgrading all of your code to use mysqli_* or PDO. See Deprecated features in PHP 5.5.x and Why are PHP's mysql_ functions deprecated? for mysql_* deprecation notes.
You should move the database connection variables to the top of your code (so they are outside of the if statement)
$db_host = 'localhost';
$db_user = 'root';
$database = 'homedb';
Related
Hey guys im trying to pass some data from an (oracle) fetch_array to a variable array and then use that array data to check if the data exists on a mysql db and create any rows that dont currently exist.. this is what i have so far.
the problem is its only checks/creates 1 entry of the array and doesn't check/created the entire array data. i think i would need to use a for loop to process all the array data concurrently
<?php
$conn = oci_connect('asdsdfsf');
$req_number = array();
if (!$conn) {
$e = oci_error();
trigger_error(htmlentities($e['message'], ENT_QUOTES), E_USER_ERROR);
}
$stid = oci_parse($conn, " SELECT WR.REQST_NO
FROM DEE_PRD.WORK_REQST WR
WHERE WR.WORK_REQST_STATUS_CD = 'PLAN' AND WR.DEPT_CD ='ISNG'
");
oci_execute($stid);
while (($row = oci_fetch_array($stid, OCI_BOTH+OCI_RETURN_NULLS)) != false) {
// Use the uppercase column names for the associative array indices
$req_number[]= $row['REQST_NO'];
}
oci_free_statement($stid);
oci_close($conn);
//MYSQL
//Connection Variables
//connect to MYSQL
$con = mysqli_connect($servername,$username,$password,$dbname);
if (!$con)
{
die('Could not connect: ' . mysqli_error());
}
// lets check if this site already exists in DB
$result = mysqli_query($con,"
SELECT EXISTS(SELECT 1 FROM wr_info WHERE REQST_NO = '$req_number') AS mycheck;
");
while($row = mysqli_fetch_array($result))
{
if ($row['mycheck'] == "0") // IF site doesnt exists lets add it to the MYSQL DB
{
$sql = "INSERT INTO wr_info (REQST_NO)VALUES ('$req_number[0]')";
if (mysqli_query($con, $sql)) {
$created = $req_number." Site Created Successfully";
} else {
echo "Error: " . $sql . "<br>" . mysqli_error($con);
}
}else{ // if site is there lets get some variables if they are present...
$result = mysqli_query($con,"
SELECT *
FROM wr_info
WHERE REQST_NO = '$req_number[0]'
");
while($row = mysqli_fetch_array($result))
{
$do some stuff
}
}
}
mysqli_close($con);
?>
You create an array:
$req_number = array();
And loop over records to assign values to the array:
while (($row = oci_fetch_array($stid, OCI_BOTH+OCI_RETURN_NULLS)) != false) {
$req_number[]= $row['REQST_NO'];
}
But then never loop over that array. Instead, you're only referencing the first record in the array:
$sql = "INSERT INTO wr_info (REQST_NO)VALUES ('$req_number[0]')";
// etc.
(Note: There are a couple of places where you directly reference the array itself ($req_number) instead of the element in the array ($req_number[0]), which is likely an error. You'll want to correct those. Also: You should be using query parameters and prepared statements. Getting used to building SQL code from concatenating values like that is a SQL injection vulnerability waiting to happen.)
Instead of just referencing the first value in the array, loop over the array. Something like this:
for($i = 0; $i < count($req_number); $i++) {
// The code which uses $req_number, but
// referencing each value: $req_number[$i]
}
so I am new in StackOverflow :), I have a problem to answer a question in php5,
This is the question :
Create a PHP 5.4 script to check availability of many Sites (via Echo Protocol):
* Get the list of sites (domains) from MySQL database.
so I write this script and I want if it's the good answer :
<?php
$dbname = 'mysql_dbname';
if (!mysql_connect('mysql_host', 'mysql_user', 'mysql_password')) {
echo 'Could not connect to mysql';
exit;
}
$sql = "SHOW TABLES FROM $dbname";
$result = mysql_query($sql);
if (!$result) {
echo "DB Error, could not list tables\n";
echo 'MySQL Error: ' . mysql_error();
exit;
}
while ($row = mysql_fetch_row($result)) {
echo "Table: {$row[0]}\n";
}
mysql_free_result($result);
?>
try something like this
$con = mysqli_connect(‘hostname’, ‘username’, ‘password’, ‘database’);
$result = mysqli_query($con , "select column_name from table_name");
while($data = mysqli_fetch_array($result))
{
echo $data['column_name'];
}
this will tell you the databases that have rights to be connected to
mysql_connect('localhost') or die ("Connect error");
$res = mysql_query("SHOW DATABASES");
while ($row = mysql_fetch_row($res)) {
echo $row[0], '<br/>';
}
You need to select a database before you use mysql_query.
$connection = mysql_connect('mysql_host', 'mysql_user', 'mysql_password');
//Write code to Check the connection
mysql_select_db('database_name', $connection);
//Write code to query
But as Robin mentioned in the comment, mysql* apis are deprecated. Use Mysqli or Pdo_Mysql. details here: http://in3.php.net/mysql_select_db
I'm learning PHP and I'm well versed with Java and C. I was given a practice assignment to create a shopping project. I need to pull out the products from my database. I'm using the product id to do this. I thought of using for loop but I can't access the prod_id from the database as a condition to check! Can anybody help me?! I have done all the form handling but I need to output the products. This is the for-loop I am using. Please let me know if I have to add any more info. Thanks in advance :)
for($i=1; $i + 1 < prod_id; $i++)
{
$query = "SELECT * FROM products where prod_id=$i";
}
I would suggest that you use PDO. This method will secure all your SQLand will keep all your connections closed and intact.
Here is an example
EXAMPLE.
This is your dbc class (dbc.php)
<?php
class dbc {
public $dbserver = 'server';
public $dbusername = 'user';
public $dbpassword = 'pass';
public $dbname = 'db';
function openDb() {
try {
$db = new PDO('mysql:host=' . $this->dbserver . ';dbname=' . $this->dbname . ';charset=utf8', '' . $this->dbusername . '', '' . $this->dbpassword . '');
} catch (PDOException $e) {
die("error, please try again");
}
return $db;
}
function getproduct($id) {
//prepared query to prevent SQL injections
$query = "SELECT * FROM products where prod_id=?";
$stmt = $this->openDb()->prepare($query);
$stmt->bindValue(1, $id, PDO::PARAM_INT);
$stmt->execute();
$rows = $stmt->fetchAll(PDO::FETCH_ASSOC);
return $rows;
}
?>
your PHP page:
<?php
require "dbc.php";
for($i=1; $i+1<prod_id; $i++)
{
$getList = $db->getproduct($i);
//for each loop will be useful Only if there are more than one records (FYI)
foreach ($getList as $key=> $row) {
echo $row['columnName'] .' key: '. $key;
}
}
First of all, you should use database access drivers to connect to your database.
Your query should not be passed to cycle. It is very rare situation, when such approach is needed. Better to use WHERE condition clause properly.
To get all rows from products table you may just ommit WHERE clause. Consider reading of manual at http://dev.mysql.com/doc.
The statement selects all rows if there is no WHERE clause.
Following example is for MySQLi driver.
// connection to MySQL:
// replace host, login, password, database with real values.
$dbms = mysqli_connect('host', 'login', 'password', 'database');
// if not connected then exit:
if($dbms->connect_errno)exit($dbms->connect_error);
$sql = "SELECT * FROM products";
// executing query:
$result = $dbms->query($sql);
// if query failed then exit:
if($dbms->errno)exit($dbms->error);
// for each result row as $product:
while($product = $row->fetch_assoc()){
// output:
var_dump($product); // replace it with requied template
}
// free result memory:
$result->free();
// close dbms connection:
$dbms->close();
for($i=1;$i+1<prod_id;$i++) {
$query = "SELECT * FROM products where prod_id=$i";
$result = mysqli_query($query, $con);
$con is the Database connection details
you can use wile loop to loop thru each rows
while ($row = mysqli_fetch_array($result))
{
......
}
}
Hope this might work as per your need..
for($i=1; $i+1<prod_id; $i++) {
$query = "SELECT * FROM products where prod_id = $i";
$result = mysql_query($query);
while ($row = mysql_fetch_array($result, MYSQL_NUM)) {
print_r($row);
}
}
I think you want all records from your table, if this is the requirement you can easily do it
$query = mysql_query("SELECT * FROM products"); // where condition is optional
while($row=mysql_fetch_array($query)){
print_r($row);
echo '<br>';
}
This will print an associative array for each row, you can access each field like
echo $row['prod_id'];
I want to be able to parse a SQL database with PHP to output an XML, however I cannot get it to show the table name and all of it's contents. I could use some help, here is my code without login information:
I have defined the db server,mdb user and pass, and db name; should anything else be defined?
<?php
// connection to the database
$dbhandle = mysql_connect(DB_SERVER, DB_USER, DB_PASS)
or die("Unable to connect to MySQL");
// select a database to work with
$selected = mysql_select_db(DB_NAME, $dbhandle)
or die("Could not select data");
// return all available tables
$result_tbl = mysql_query( "SHOW TABLES FROM "DB_NAME, $dbhandle );
$tables = array();
while ($row = mysql_fetch_row($result_tbl)) {
$tables[] = $row[0];
}
$output = "<?xml version=\"1.0\" ?>\n";
$output .= "<schema>";
// iterate over each table and return the fields for each table
foreach ( $tables as $table ) {
$output .= "<table name=\"$table\">";
$result_fld = mysql_query( "SHOW FIELDS FROM "$table, $dbhandle );
while( $row1 = mysql_fetch_row($result_fld) ) {
$output .= "<field name=\"$row1[0]\" type=\"$row1[1]\"";
$output .= ($row1[3] == "PRI") ? " primary_key=\"yes\" />" : " />";
}
$output .= "</table>";
}
$output .= "</schema>";
// tell the browser what kind of file is come in
header("Content-type: text/xml");
// print out XML that describes the schema
echo $output;
// close the connection
mysql_close($dbhandle);
?>
I think it's better to use standard php class XmlWriter for this one.
Look at http://www.php.net/manual/en/book.xmlwriter.php
$result_tbl = mysql_query( "SHOW TABLES FROM "DB_NAME, $dbhandle );
^^^---typo
$result_fld = mysql_query( "SHOW FIELDS FROM "$table, $dbhandle );
^^^---typo
You've got at least two typos in your queries. Most likely you want this:
$result_tbl = mysql_query("SHOW TABLES FROM " . DB_NAME, $dbhandle) or die(mysql_error());
and
$result_fld = mysql_query("SHOW FIELDS FROM $table", $dbhandle) or die(mysql_error());
Note the concatenation operator (.) on the first one, and the addition of or die(...). Never assume a query succeeds. even if the query string itself is syntactically correct, there's far too many OTHER reasons is could fail to NOT check for an error condition.
I was wondering if there's a way in PHP to list all available databases by usage of mysqli. The following works smooth in MySQL (see php docs):
$link = mysql_connect('localhost', 'mysql_user', 'mysql_password');
$db_list = mysql_list_dbs($link);
while ($row = mysql_fetch_object($db_list)) {
echo $row->Database . "\n";
}
Can I Change:
$db_list = mysql_list_dbs($link); // mysql
Into something like:
$db_list = mysqli_list_dbs($link); // mysqli
If this is not working, would it be possible to convert a created mysqli connection into a regular mysql and continue fetching/querying on the new converted connection?
It doesn't appear as though there's a function available to do this, but you can execute a show databases; query and the rows returned will be the databases available.
EXAMPLE:
Replace this:
$db_list = mysql_list_dbs($link); //mysql
With this:
$db_list = mysqli_query($link, "SHOW DATABASES"); //mysqli
I realize this is an old thread but, searching the 'net still doesn't seem to help. Here's my solution;
$sql="SHOW DATABASES";
$link = mysqli_connect($dbhost,$dbuser,$dbpass) or die ('Error connecting to mysql: ' . mysqli_error($link).'\r\n');
if (!($result=mysqli_query($link,$sql))) {
printf("Error: %s\n", mysqli_error($link));
}
while( $row = mysqli_fetch_row( $result ) ){
if (($row[0]!="information_schema") && ($row[0]!="mysql")) {
echo $row[0]."\r\n";
}
}
Similar to Rick's answer, but this is the way to do it if you prefer to use mysqli in object-orientated fashion:
$mysqli = ... // This object is my equivalent of Rick's $link object.
$sql = "SHOW DATABASES";
$result = $mysqli->query($sql);
if ($result === false) {
throw new Exception("Could not execute query: " . $mysqli->error);
}
$db_names = array();
while($row = $result->fetch_array(MYSQLI_NUM)) { // for each row of the resultset
$db_names[] = $row[0]; // Add db name to $db_names array
}
echo "Database names: " . PHP_EOL . print_r($db_names, TRUE); // display array
Here is a complete and extended solution for the answer, there are some databases that you do not need to read because those databases are system databases and we do not want them to appear on our result set, these system databases differ by the setup you have in your SQL so this solution will help in any kind of situations.
first you have to make database connection in OOP
//error reporting Procedural way
//mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);
//error reporting OOP way
$driver = new mysqli_driver();
$driver->report_mode = MYSQLI_REPORT_ALL & MYSQLI_REPORT_STRICT;
$conn = new mysqli("localhost","root","kasun12345");
using Index array of search result
$dbtoSkip = array("information_schema","mysql","performance_schema","sys");
$result = $conn->query("show databases");
while($row = $result->fetch_array(MYSQLI_NUM)){
$print = true;
foreach($dbtoSkip as $key=>$vlue){
if($row[0] == $vlue) {
$print=false;
unset($dbtoSkip[$key]);
}
}
if($print){
echo '<br/>'.$row[0];
}
}
same with Assoc array of search result
$dbtoSkip = array("information_schema","mysql","performance_schema","sys");
$result = $conn->query("show databases");
while($row = $result->fetch_array(MYSQLI_ASSOC)){
$print = true;
foreach($dbtoSkip as $key=>$vlue){
if($row["Database"] == $vlue) {
$print=false;
unset($dbtoSkip[$key]);
}
}
if($print){
echo '<br/>'.$row["Database"];
}
}
same using object of search result
$dbtoSkip = array("information_schema","mysql","performance_schema","sys");
$result = $conn->query("show databases");
while($obj = $result->fetch_object()){
$print = true;
foreach($dbtoSkip as $key=>$vlue){
if( $obj->Database == $vlue) {
$print=false;
unset($dbtoSkip[$key]);
}
}
if($print){
echo '<br/>'. $obj->Database;
}
}