$result = mysql_query("SELECT * FROM MasjidMaster WHERE MasjidName LIKE ('%moh%')") or die mysql_error();
The error i get is
Parse error: syntax error, unexpected T_STRING in /home/maximtec/public_html/masjid_folder/MasjidFinderScripts/find_by_name.php on line 24
This query does work when i use it in MySQL but it doesn't when I place it in a PHP Script
Please suggest a solution
------------EDIT :After changing query from the received answers-------------------------------------
Well I updated my query but now I am getting null results
{"masjids":[{"MasjidName":null,"Address":null,"Latitude":null,"Longitude":null}],"success":1,"masjid":[]}
Following is my full script :
<?php
/*
* Following code will get single product details
* A product is identified by product id (pid)
*/
// array for JSON response
$response = array();
// include db connect class
//require_once __DIR__ . '/db_connect.php';
require_once dirname(__FILE__ ). '/db_connect.php';;
// connecting to db
$db = new DB_CONNECT();
// check for post data
if (isset($_GET["MasjidName"])) {
$MasjidName = $_GET['MasjidName'];
// get a product from products table
$result = mysql_query("SELECT * FROM `MasjidMaster` WHERE `MasjidName` LIKE '%moh%'") or die(mysql_error());
$response["masjids"] = array();
if (!empty($result)) {
// check for empty result
if (mysql_num_rows($result) > 0) {
while ($row = mysql_fetch_array($result)) {
$row = mysql_fetch_array($result);
$masjid = array();
$masjid["MasjidName"] = $row["MasjidName"];
$masjid["Address"] = $row["Address"];
$masjid["Latitude"] = $row["Latitude"];
$masjid["Longitude"] = $row["Longitude"];
// success
$response["success"] = 1;
// user node
$response["masjid"] = array();
array_push($response["masjids"], $masjid);
// array_push($response["masjid"], $masjid);
}
// echoing JSON response
echo json_encode($response);
} else {
// no product found
$response["success"] = 0;
$response["message"] = "No product found";
// echo no users JSON
echo json_encode($response);
}
} else {
// no product found
$response["success"] = 0;
$response["message"] = "No product found";
// echo no users JSON
echo json_encode($response);
}
} else {
// required field is missing
$response["success"] = 0;
$response["message"] = "Required field(s) is missing";
// echoing JSON response
echo json_encode($response);
}
?>
$result = mysql_query("SELECT * FROM `MasjidMaster` WHERE `MasjidName` LIKE '%moh%'") or die(mysql_error());
A little more tweaking, for good practice wrap table names and table columns in ``.
You also shouldn't need () around ('%moh%')
Try this-
$result = mysql_query("SELECT * FROM MasjidMaster WHERE MasjidName LIKE ('%moh%')") or die(mysql_error());
You forget parentheses with die(mysql_error())
try this:
$result = mysql_query("SELECT * FROM MasjidMaster WHERE MasjidName LIKE '%moh%' ") or die(mysql_error());
use die() like this die(mysql_error()) .
Related
I'm trying to convert an Array to JSON in PHP I have two identical pieces of code except for them referencing to different databases and data. could someone help me?
The code that works:
get_all_products.php
<?php
/*
* Following code will list all the products
*/
// array for JSON response
$response = array();
// include db connect class
require_once __DIR__ . '/db_connect.php';
// connecting to db
$db = new DB_CONNECT();
// get all products from products table
$result = mysql_query("SELECT * FROM products") or die(mysql_error());
// check for empty result
if (mysql_num_rows($result) > 0) {
// looping through all results
// products node
$response["products"] = array();
while ($row = mysql_fetch_array($result)) {
// temp product array
$product = array();
$product["pid"] = $row["pid"];
$product["pname"] = $row["pname"];
$product["pprice"] = $row["pprice"];
$product["pamount"] = $row["pamount"];
// push single product into final response array
array_push($response["products"], $product);
}
// success
$response["success"] = 1;
// echoing JSON response
echo json_encode($response);
} else {
// no products found
$response["success"] = 0;
$response["message"] = "No products found";
// echo no users JSON
echo json_encode($response);
}
?>
The one that doesn't work:
post2.php
<?php
/*
* Following code will list all the users
*/
// array for JSON response
$response = array();
// include db connect class
require_once __DIR__ . '/db_connect.php';
// connecting to db
$db = new DB_CONNECT();
// get all users from users table
$result = mysql_query("SELECT * FROM users") or die(mysql_error());
// check for empty result
if (mysql_num_rows($result) > 0) {
// looping through all results
// users node
$response["users"] = array();
while ($row = mysql_fetch_array($result)) {
// temp user array
$user = array();
$user["ID"] = $row["ID"];
$user["name"] = $row["name"];
$user["balance"] = $row["balance"];
$user["pin"] = $row["pin"];
// push single user into final response array
array_push($response["users"], $user);
}
// success
$response["success"] = 1;
// echoing JSON response
echo json_encode($response);
} else {
// no users found
$response["success"] = 0;
$response["message"] = "No users found";
// echo no users JSON
echo json_encode($response);
}
?>
I Asked my teacher too but he didn't know the answer either.
I am getting null results when I use the following Like query in PHP Script
$MasjidName = $_GET['MasjidName'];
$Percent = "%";
$search = $Percent.$MasjidName.$Percent;
echo $search;
$sql = "SELECT * FROM `MasjidMaster` WHERE `MasjidName` LIKE '".$search."'";
// get a product from products table
$result = mysql_query($sql) or die(mysql_error());
I have tried the following too
$result = mysql_query("SELECT * FROM `MasjidMaster` WHERE `MasjidName` LIKE '%moh%'") or die(mysql_error());
The following is the null result I have been getting
{"masjids":[{"MasjidName":null,"Address":null,"Latitude":null,"Longitude":null}],"success":1,"masjid":[]}
whole code added below the following is the script i have been trying to get work
<?php
$response = array();
require_once dirname(__FILE__ ). '/db_connect.php';;
$db = new DB_CONNECT();
if (isset($_GET["MasjidName"]))
{
$MasjidName = $_GET['MasjidName'];
$MasjidName = mysql_real_escape_string($MasjidName); // you have to escape your variable here.
$sql = "SELECT * FROM `MasjidMaster` WHERE `MasjidName` LIKE '%$MasjidName%'";
$result = mysql_query($sql) or die(mysql_error());
$response["masjids"] = array();
if (!empty($result)) {
// check for empty result
if (mysql_num_rows($result) > 0) {
while ($row = mysql_fetch_array($result)) {
$row = mysql_fetch_array($result);
$masjid = array();
$masjid["MasjidName"] = $row["MasjidName"];
$masjid["Address"] = $row["Address"];
$masjid["Latitude"] = $row["Latitude"];
$masjid["Longitude"] = $row["Longitude"];
// success
$response["success"] = 1;
// user node
$response["masjid"] = array();
array_push($response["masjids"], $masjid);
}
// echoing JSON response
echo json_encode($response);
} else {
// no product found
$response["success"] = 0;
$response["message"] = "No product found";
// echo no users JSON
echo json_encode($response);
}
} else {
// no product found
$response["success"] = 0;
$response["message"] = "No product found";
// echo no users JSON
echo json_encode($response);
}
} else {
// required field is missing
$response["success"] = 0;
$response["message"] = "Required field(s) is missing";
// echoing JSON response
echo json_encode($response);
}
?>
Try this:
$MasjidName = $_GET['MasjidName'];
$MasjidName = mysql_real_escape_string($MasjidName); // you have to escape your variable here.
$sql = "SELECT * FROM `MasjidMaster` WHERE `MasjidName` LIKE '%$MasjidName%'";
$result = mysql_query($sql) or die(mysql_error());
Try using
$sql = "SELECT * FROM MasjidMaster WHERE MasjidName LIKE '".$search."'";
I tried SELECT * FROM Customers WHERE City LIKE 's%'; and it gave me perfectly fine result. But when i tried SELECT * FROM 'Customers' WHERE 'City' LIKE 's%'; it gave me a null result.
Just remove '' and give it a try. Hope it helps.
I have a problem with my PHP query. I just need to get the prescription for the clicked patient in listview.
Here is the table structure for prescription.
I need patient with id 79 to display all his corresponding prescription. Please help me.
Here is the PHP.
<?php
/*
* Following code will list all the products
*/
// array for JSON response
$response = array();
// include db connect class
require_once __DIR__ . '/db_connect.php';
// connecting to db
$db = new DB_CONNECT();
if (isset($_GET["uid"])) {
$uid = $_GET['uid'];
// get all products from products table
$result = mysql_query("select * from prescription") or die(mysql_error());
// check for empty result
if (mysql_num_rows($result) > 0) {
// looping through all results
// products node
$response["prescription"] = array();
while ($row = mysql_fetch_array($result)) {
// temp user array
$prescription = array();
$prescription["pres_id"] = $row["pres_id"];
$prescription["pres_name"] = $row["pres_name"];
//$product["price"] = $row["price"];
// push single product into final response array
array_push($response["prescription"], $prescription);
}
// success
$response["success"] = 1;
// echoing JSON response
echo json_encode($response);
} else {
// no products found
$response["success"] = 0;
$response["message"] = "No prescriptions found";
// echo no users JSON
echo json_encode($response);
}
}else {
$response["success"] = 0;
$response["message"] = "Required field(s) is missing";
// echoing JSON response
echo json_encode($response);
}
}
?>
To get all the prescriptions for a selected patient, wouldn't it be:
mysql_query("select * from prescription where patient_user_fkey = '$patientID'")or die(mysql_error());
instead of:
mysql_query("select * from prescription") or die(mysql_error());
I have this code passing a variable from a url. When i use $_GET method it returns me a json with no products found but when i give manually the the value that $user_email has from the url it returns me the correct json! what is wrong and how can i correct it? thank you
URL: http://***********/android_connect/get_all_products.php?user_email=m
<?php
/*
* Following code will list all the products
*/
// array for JSON response
$response = array();
$user_email= $_GET['user_email'];
// include db connect class
require_once __DIR__ . '/db_connect.php';
// connecting to db
$db = new DB_CONNECT();
// get all products from products table
$test= "SELECT *FROM products WHERE user_email= '" .$user_email. "'";
//echo $test;
$result = mysql_query($test) or die(mysql_error());
// check for empty result
if (mysql_num_rows($result) > 0) {
// looping through all results
// products node
$response["products"] = array();
while ($row = mysql_fetch_array($result)) {
// temp user array
$product = array();
$product["pid"] = $row["pid"];
$product["firstname"] = $row["firstname"];
$product["lastname"] = $row["lastname"];
$product["email"] = $row["email"];
$product["phone"] = $row["phone"];
$product["address"] = $row["address"];
$product["created_at"] = $row["created_at"];
$product["updated_at"] = $row["updated_at"];
$product["user_email"] = $row["user_email"];
// push single product into final response array
array_push($response["products"], $product);
}
// success
$response["success"] = 1;
// echoing JSON response
echo json_encode($response);
} else {
// no products found
$response["success"] = 0;
$response["message"] = "No products found";
// echo no users JSON
echo json_encode($response);
}
?>
Try this:
$test= "SELECT * FROM products WHERE user_email = '$user_email'";
EDIT:
$test= "SELECT * FROM products WHERE user_email = $user_email";
i'm having a problem where for instance: we have 2 users, so user A insert the data into the database and it can be retrieved with no problem. User B insert the data into the database and when user B want to retrieve the data, the user B gets the data of user A not user B.
What will be causing that problem?
Here my php code for getting data from mysql database:
<?php
// array for JSON response
$response = array();
// include db connect class
require_once __DIR__ . '/db_connect.php';
// connecting to db
$db = new DB_CONNECT();
// check for post data
if (isset($_GET["subject_id"])) {
$subject_id = $_GET['subject_id'];
// get a product from products table
$result = mysql_query("SELECT * FROM subject_offered WHERE subject_id = $subject_id");
if (!empty($result)) {
// check for empty result
if (mysql_num_rows($result) > 0) {
$result = mysql_fetch_array($result);
$product = array();
$product["subject_id"] = $result["subject_id"];
$product["lecturer_name"] = $result["lecturer_name"];
$product["time_offered"] = $result["time_offered"];
$product["subject_details"] = $result["subject_details"];
//$product["updated_at"] = $result["updated_at"];
// success
$response["success"] = 1;
// user node
$response["product"] = array();
array_push($response["product"], $product);
// echoing JSON response
echo json_encode($response);
} else {
// no product found
$response["success"] = 0;
$response["message"] = "No subject found";
// echo no users JSON
echo json_encode($response);
}
} else {
// no product found
$response["success"] = 0;
$response["message"] = "No subject found";
// echo no users JSON
echo json_encode($response);
}
} else {
// required field is missing
$response["success"] = 0;
$response["message"] = "Required field(s) is missing";
// echoing JSON response
echo json_encode($response);
}
?>
// get a product from products table
$result = mysql_query("SELECT * FROM subject_offered WHERE subject_id = $subject_id");
should be
// get a product from products table
$result = mysql_query("SELECT * FROM subject_offered WHERE subject_id = '$subject_id' ");
//some times when you miss '' this single commas then it also create a problem