php- cant get result from mysql (mysqli) - php

I'm attempting to set up an API for an app project.
I've got a mysql table called 'users', which I've added a row to.
using the code:
// Create connection
$mysqli = new mysqli("localhost","user", "pass", "db");
// Check connection
if($mysqli->connect_errno){
$result = "Failed to connect to MySQL: " . mysqli_connect_error();
print_r( json_encode($result) );
return false;
}
$row = $mysqli->query("SELECT * FROM users");
print_r( json_encode($row) );
I get an empty result, how come? (connection doesn't throw an error)
to be exact i get:
{"current_field":null,"field_count":null,"lengths":null,"num_rows":null,"type":null}
EDIT:
got the answer to ym original question, thanks!
so now using the code:
$row = $mysqli->query("SELECT * FROM users WHERE email = '".$email."'");
$result = $row->fetch_array();
print_r( json_encode($result) );
I get the result:
{"0":"test","username":"test","1":"test#test.com","email":"test#test.com","2":"test","password":"test","3":"2013-10-18 22:22:53","date_registered":"2013-10-18 22:22:53","4":"1","id":"1"}
where what i want is something like:
{"username":"test","password":"test","email":"test#test.com", ...etc }
how do i get that?

Try this:
$result = $mysqli->query("SELECT * FROM users");
$row = $result->fetch_array(MYSQLI_ASSOC);
print json_encode($row); // json_encode returns a string...
Try this for your associative array:
while($row = $result->fetch_array(MYSQLI_ASSOC))
{
$rows[] = $row;
}
print json_encode($rows);
or you can try... $rows = $result->fetch_all(MYSQLI_ASSOC);

mysqli_query() will return a mysqli_result, you need to fetch (as an array in this case) your rows before doing anything with them. Added a line:
// Create connection
$mysqli = new mysqli("localhost","user", "pass", "db");
// Check connection
if($mysqli->connect_errno){
$result = "Failed to connect to MySQL: " . mysqli_connect_error();
print_r( json_encode($result) );
return false;
}
// Get a mysql_result
$row = $mysqli->query("SELECT * FROM users");
// Get it into an array without numeric indexes
$result = $row->fetch_assoc();
// Display the row
print_r( json_encode($result) );

Related

Error is Trying to get property 'id' of non-object, but seems object has

I am using PHP for 1st time.
I am just trying to fetch data from database. Here is my code -
try{
///try to connect with database
$conn=new PDO("mysql:host=localhost:3306;dbname=try", "root", "abcdef12");
$conn->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
}
catch(PDOException $ex){
echo "error";
}
$mysqlcode="SELECT * FROM users";
$ret=$conn->query($mysqlcode);
foreach($ret as $ret1)
print $ret1->id;
in the browser,
showing this error:
Notice: Trying to get property 'id' of non-object in /opt/lampp/htdocs/secAsite/verifylogin.php on line 44
In the users table, I have 4 data.
In here, why $ret seems non object. I can't find anything wrong. How to fetch data. Or debug the object.
I have checked your code. Everything is fine. Just do this.
foreach($ret as $ret1)
print $ret1['id'];
You are not getting data because you are not fetching it..
you have to fetch the data before print it
updated code
$mysqlcode = "SELECT * FROM users";
$ret = $conn->query($mysqlcode);
if ($ret->num_rows > 0) {
// output data of each row
while($row = $ret->fetch_assoc()) {
echo "id: " . $ret["id"] . "<br>";
}
} else {
echo "0 results";
}
$conn->close();
that'll work fine.
Before trying to use Object Oriented PHP, is much easier to understand Procedural. Take a look at the code below.
First, connect to database:
<?php
//Store log in info into variables
$servername = 'localhost';
$serveruser = 'youruser';
$serverpassword = 'yourpassword';
$serverdatabase = 'yourdatabase';
// Create connection
$conn = mysqli_connect( $servername, $serveruser, $serverpassword, $serverdatabase );
if ( !$conn ) {echo "Connection Fail" . mysqli_connect_error();}
//else {echo "Connected to database $serverdatabase";}
?>
Then, call the information from your database and your table and store it into an associative array $data
$sql = "SELECT * FROM `table`";
$result = mysqli_query($conn, $sql);
$data = mysqli_fetch_all($result, MYSQLI_ASOCC);
mysqli_free_result($result);
Now you can use the array to print the results with print_r
echo '<pre>'; print_r($data); echo '</pre>';}
Now you are ready to use foreach depending on your table columns,for example if you have ID column then
<?php foreach ($data as $value){
echo $value['id]."<br>";
} ?>
It will print out all the IDs. Hope this helps.

json_encode and json_decode in mySql misunderstanding

I am stuck in my PHP page for a json_encode since a couple of days after having done 9999 tests.
The goal is to log on a swift app using mySql database.
So I guess i have something wrong in my query, and i dont know how to send answer of found/not found user in json_encode().
Here is my php code:
<?php
$json = file_get_contents('php://input');
$obj = json_decode($json);
// i get the email and password sent as parameter
// my swift code is working, and the next line is ok.
$logs[] = array('email' => $obj->email, 'pass' => $obj->pass);
//echo json_encode($logs); // to send back my logs (ok)
// Create connection
$con = mysqli_connect("localhost", "xx", "xx", "xx");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
//echo json_encode($logs);
// its working until here if i execute the previous line and put the rest in comment
// from the next line its making errors such :
// Error Domain=NSCocoaErrorDomain Code=3840 "JSON text did not start with array or object and option to allow fragments not set."
$sql = "SELECT * FROM clients WHERE email = '".$logs['email']."'";
//echo json_encode($logs); //not working at this place if i put the rest in comment.
// Check if there are results
if ($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);
}
//check the corresponding password
if ( $logs['pass'] == $resultArray[4]) {
// i m not sure how to send the answer as a json
echo json_encode(true);
} else {
echo json_encode(false);
}
}
// Close connections
mysqli_close($con);
?>
If I try in php without all json stuff, its working since I put
$sql = "SELECT * FROM clients WHERE email = 'ben#test.be'";
as a test.
The following code works:
<?php
$json = file_get_contents('php://input');
$obj = json_decode($json);
$logs[] = array('email' => $obj->email, 'pass' => $obj->pass);
$con = mysqli_connect("localhost", "xx", "xx", "xx");
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$sql = "SELECT * FROM clients";
if ($result = mysqli_query($con, $sql))
{
$emparray = array();
while($row =mysqli_fetch_assoc($result))
{
$emparray[] = $row;
}
echo json_encode($emparray);
}
mysqli_close($con);
?>
But if I put $sql = "SELECT * FROM clients WHERE email = '$logs['email']'"; instead of $sql = "SELECT * FROM clients";
Its not working anymore. It doesn't like my conditional query. What did I misunderstood?

Flattening a SQL query result for PHP array

I have a SQL table (modules) with two columns (id, name). Now I can retrieve the rows from this through a PHP script but what I want is to use the value of id as the key, and the value of name as the value, in a multidimensional array. Then I want to be able to encode those into a JSON, retaining the relationship between key/value. I've muddled something together but it returns null.
the relevant code from index.php
$mod1 = $core["module1"];
$mod2 = $core["module2"];
$modules = $db->getModulesById($mod1, $mod2); //module names & ids
$response["module"]["mod1"] = $modules[$mod1];
$response["module"]["mod2"] = $modules[$mod2];
$response["module"]["mod1name"] = $modules[$mod1]["name"];
$response["module"]["mod2name"] = $modules[$mod2]["name"];
echo json_encode($response);
The function from DB_Functions.php
public function getModulesById($mod1, $mod2) {
require_once 'include/Config.php';
$con = mysqli_connect(DB_HOST, DB_USER, DB_PASSWORD);
// Check connection
if (!$con)
{
die("Connection error: " . mysqli_connect_error());
}
// selecting database
mysqli_select_db($con, DB_DATABASE) or die(mysqli_connect_error());
$query = "SELECT * FROM modules WHERE id= '$mod1' OR id='$mod2'";
$result = mysqli_query($con, $query);
$arr = array();
while($row = mysqli_fetch_assoc($result)) {
// process each row
//each element of $arr now holds an id and name
$arr[] = $row;
}
// return user details
return mysqli_fetch_array($arr);
close();
}
I've looked around but I'm just not 'getting' how the query return is then broken down into key/value for a new array. If someone could ELI5 I'd appreciate it. I'm just concerned with this aspect, it's a personal project so I'm not focusing on security issues as yet, thanks.
You are pretty well there
public function getModulesById($mod1, $mod2) {
require_once 'include/Config.php';
$con = mysqli_connect(DB_HOST, DB_USER, DB_PASSWORD, DB_DATABASE);
// Check connection
if (!$con) {
die("Connection error: " . mysqli_connect_error());
}
$query = "SELECT * FROM modules WHERE id= '$mod1' OR id='$mod2'";
$result = mysqli_query($con, $query);
$arr = array();
while($row = mysqli_fetch_assoc($result)) {
$arr[] = $row;
}
// here is wrong
//return mysqli_fetch_array($arr);
// instead return the array youy created
return $arr;
}
And call it and then just json_encode the returned array
$mod1 = $core["module1"];
$mod2 = $core["module2"];
$modules = $db->getModulesById($mod1, $mod2); //module names & ids
$response['modules'] = $modules;
echo json_encode($response);
You should really be using prepared and paramterised queries to avoid SQL Injection like this
public function getModulesById($mod1, $mod2) {
require_once 'include/Config.php';
$con = mysqli_connect(DB_HOST, DB_USER, DB_PASSWORD, DB_DATABASE);
// Check connection
if (!$con) {
die("Connection error: " . mysqli_connect_error());
}
$sql = "SELECT * FROM modules WHERE id= ? OR id=?";
$stmt = $con->prepare($sql);
$stmt->bind_param('ii', $mod1, $mod2);
$stmt->execute();
$result = $stmt->get_result();
$arr = array();
while($row = mysqli_fetch_assoc($result)) {
$arr[] = $row;
}
// here is wrong
//return mysqli_fetch_array($arr);
// instead return the array youy created
return $arr;
}
mysqli_fetch_array requires the result of a mysqli_query result. Passing the constructed array to mysqli_fetch_array() is not going to work.
If you want to have a specific value from a row to use as its key, you can't resolve this with any mysqli_* function. You could however construct it yourself:
while($row = mysqli_fetch_assoc($result)) {
// process each row
//each element of $arr now holds an id and name
$arr[$row['id']] = $row;
}
mysqli_close($con);
return $arr;
You should close the connection before returning the result, code positioned after a return will not be executed.

Im trying to encode all data from a table in JSON but there is no result

Here is the webpage in question
http://liamure.xyz/rsk/getdata.php
Here is the database
Database
And here is the SQL code
<?php
$con=mysqli_connect('host', 'user', 'Password', 'database');
if(mysqli_connect_errno())
{
echo "Failed Connection" . mysqli_connect_errno();
}
else
{
$sth = mysqli_query("SELECT * FROM quotes");
$rows = array();
while($r = mysqli_fetch_assoc($sth)) {
echo"1";
$rows[] = $r;
}
print json_encode($rows);
}
?>
When the webpage is run all that gets returned is "[]" (You can see this by clicking the URL above)
The first argument of mysqli_query function is the connection link object:
$sth = mysqli_query($con, "SELECT * FROM quotes");

Using PHP to query a MDB file, and return JSON

I have a Microsoft Access Database, and I am trying to query the table using PHP, and output valid JSON. I have an equivalent code for a MSSQL database, am I am trying to make my code do the same thing, but just for the Access database.
Here is the MSSQL code
$myServer = "server";
$myDB = "db";
$conn = sqlsrv_connect ($myServer, array('Database'=>$myDB));
$sql = "SELECT *
FROM db.dbo.table";
$data = sqlsrv_query ($conn, $sql);
$result = array();
do {
while ($row = sqlsrv_fetch_array ($data, SQLSRV_FETCH_ASSOC)) {
$result[] = $row;
}
} while (sqlsrv_next_result($data));
$json = json_encode ($result);
sqlsrv_free_stmt ($data);
sqlsrv_close ($conn);
Here is what I tried for the MDB file
$dbName = "/filename.mdb";
if (!file_exists($dbName)) {
die("Could not find database file.");
}
$db = odbc_connect("Driver={Microsoft Access Driver (*.mdb)};Dbq=$dbName", $user, $password);
$sql = "SELECT *
FROM cemetery";
$data = $db->query($sql); // I'm getting an error here
$result = array();
// Not sure what do do for this part...
do {
while ($row = fetch($data, SQLSRV_FETCH_ASSOC)) {
$result[] = $row;
}
} while (sqlsrv_next_result($data));
$json = json_encode ($result);
I kind of followed this to try to connect to the database: http://phpmaster.com/using-an-access-database-with-php/
Currently this is giving me a 500 Internal Server Error. I'm expecting a string such as this to be saved in the variable $json
[
{
"col1":"col value",
"col2":"col value",
"col3":"col value",
},
{
"col1":"col value",
"col2":"col value",
"col3":"col value",
},
{
etc...
}
]
Can someone help me port the MSSQL code I have above so I can use it with an MDB database? Thanks for the help!
EDIT: I'm commenting out the lines one by one, and it throws me the 500 error at the line $data = $db->query($sql);. I looked in the error log, and I'm getting the error Call to a member function query() on a non-object. I already have the line extension=php_pdo_odbc.dll uncommented in my php.ini file. Anyone know what the problem could be?
You only need 1 loop,
fetchAll is your iterable friend:
while ($row = $data->fetchAll(SQLSRV_FETCH_ASSOC)) {
$result[] = $row;
}
odbc_connect doesn't return an object, it returns a resource. see (http://php.net/manual/en/function.odbc-connect.php) so you would need to do something like this.
$db = odbc_connect("Driver={Microsoft Access Driver (*.mdb)};Dbq=$dbName", $user, $password);
$oexec = obdc_exec($db,$sql);
$result[] = odbc_fetch_array($oexec);
and then you can iterate over results..
see also:
http://www.php.net/manual/en/function.odbc-fetch-array.php
http://www.php.net/manual/en/function.odbc-exec.php
I finally figured it out.
<?php
// Location of database. For some reason I could only get it to work in
// the same location as the site. It's probably an easy fix though
$dbName = "dbName.mdb";
$tName = "table";
// Throws an error if the database cannot be found
if (!file_exists($dbName)) {
die("Could not find database file.");
}
// Connects to the database
// Assumes there is no username or password
$conn = odbc_connect("Driver={Microsoft Access Driver (*.mdb)};Dbq=$dbName", '', '');
// This is the query
// You have to have each column you select in the format tableName.[ColumnName]
$sql = "SELECT $tName.[ColumnOne], $tName.[ColumnTwo], etc...
FROM $dbName.$tName";
// Runs the query above in the table
$rs = odbc_exec($conn, $sql);
// This message is displayed if the query has an error in it
if (!$rs) {
exit("There is an error in the SQL!");
}
$data = array();
$i = 0;
// Grabs all the rows, saves it in $data
while( $row = odbc_fetch_array($rs) ) {
$data[$i] = $row;
$i++;
}
odbc_close($conn); // Closes the connection
$json = json_encode($data); // Generates the JSON, saves it in a variable
?>
I use this code to get results from an ODBC query into a JSON array:
$response = null;
$conn = null;
try {
$odbc_name = 'myODBC'; //<-ODBC connectyion name as is in the Windows "Data Sources (ODBC) administrator"
$sql_query = "SELECT * FROM table;";
$conn = odbc_connect($odbc_name, 'user', 'pass');
$result = odbc_exec($conn, $sql_query);
//this will show all results:
//echo odbc_result_all($result);
//this will fetch row by row and allows to change column name, format, etc:
while( $row = odbc_fetch_array($result) ) {
$json['cod_sistema'] = $row['cod_sistema'];
$json['sistema'] = $row['sistema'];
$json['cod_subsistema'] = $row['cod_subsistema'];
$json['sub_sistema'] = $row['sub_sistema'];
$json['cod_funcion'] = $row['cod_funcion'];
$json['funcion'] = $row['funcion'];
$json['func_desc_abrev'] = $row['desc_abreviada'];
$json['cod_tipo_funcion'] = $row['cod_tipo_funcion'];
$response[] = array('funcionalidad' => $json);
}
odbc_free_result($result); //<- Release used resources
} catch (Exception $e) {
$response = array('resultado' => 'err', 'detalle' => $e->getMessage());
echo 'ERROR: ', $e->getMessage(), "\n";
}
odbc_close($conn);
return $response;
And finnally encoding the response in JSON format:
echo json_encode($response);

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