I am sure that this question has been asked before, but I am unable to come up with the proper keywords (especially in english).
I am using PHP and I am trying to for loop through a parameter of a function. So the function should be called, store the retrieved data in some variables and these variables should then be inserted into a database.
However, the loops only runs once! If I substitute $id with any number it works fine, but only once.
This is a simplified version of my code:
for ($i=0; $i<9; $i++) {
$id = $rows[$i][1];
$values = getDetails($id); // This function (from another file) returns an array
$title = $values["Title"];
$year = $values["Year"];
$query= " INSERT INTO database
VALUES ('','$title','$year')";
$result = $mysqli->query($query);
}
* EDIT This is part of the getDetails function:
function getDetails($id) {
$url = "http://www.something.de/". $id . "/";
$html = file_get_html ( $url );
$title = $html->find('span[itemprop=name]');
$title = explode('>',$title[0]);
$title = explode('</span',$title[1]);
... // This might look weird and is definatly not perfect, but it works :)
$details = array("Title" => $title[0], "Year" => $year[1]);
return $details;
}
* EDIT
WOW! I found the reason ... I had a function within my function which was never used. I just commented it out and my code works just fine. I assume it is not a good idea to so anyways.
I think your $query is wrong.
Change this:
$query= " INSERT INTO database
VALUES ('','$title','$year')";
To something like this:
$query= " INSERT INTO database (field1,field2,field3)
VALUES ('','$title','$year')";
Is your ID field autoincrementing? If so you do not need the "field1" entry at all.
Happy Coding!
I had this problem also.
I could print to a table without a problem the parameters I was feeding into a function in a loop. But the function calls in the loops would only call once.
SOLUTION: Remove the location redirects and the exit(); from the function.
Hope this helps someone else.
Related
New to OOP so I am trying to figure out best practice. This code is based off an existing script I am adding to.
Most of the threads with this question tell the poster to code as so:
function ($arg1, $arg2){
//some code
}
and call:
function($a1, $a2);
I have an OOP-based function (that works) but it doesn't quite look right and when I try to call it as the suggested method, I get:
Array to string conversion .... on line .. Array
Here's my (working) function that gathers the output:
public function getMail($type, $id = 0) {
$query = $this->db->query("SELECT * FROM km_mail WHERE id = '" . (int)$id . "' AND `type` = '" . $this->db->escape($type) . "'");
foreach ($query->rows as $result) {
$mail_data[$result['title']] = $result['value'];
}
return $mail_data;
}
This is the working (but ugly) part - this returns the database column requested (but looks wrong?):
$this->model_setting_mail->getMail('order')['update_link'];
When I try to request the column like so, the array to string conversion error occurs:
$this->model_setting_mail->getMail('order','update_link');
In my example, order = $type, update_link = $result['value'] and $id = 0 is default, unless an $id is passed.
The first example you show is a shorthand way of selecting an array element from value returned by a function.
$this->model_setting_mail->getMail('order')['update_link'];
Is the same as:
$result = $this->model_setting_mail->getMail('order');
print $result['update_link'];
Second example is passing two values to a function.
They are completely different.
I have a function to process info from a database. This is called multiple times in a page. And I don't want to query the database every time. So I put the query outside. If I do that, the function doesn't work. I know this can be done because, there was a similar question somewhere in SO. But that addressed a different situation. I don't know what is wrong here. Any help will be greatly appreciated.
If I put all this code into a separate test file including the conn file and query, it works. But in my main page, where I have the functions.php included first, then conn.php and then the query and then the display code called by js fadein event, the $result refuses to work inside the function
EDIT : This code has been cleaned up as per comments received (globals replaced with variables passed to the function and variable names rationalised)
function total($item,$result,$val){
global $totRate;
while($getRates=$result->fetch_assoc()){
$gotItem= strtolower(preg_replace('/[^(\x20-\x7F)]*/',"",$getRates['item']));
$gotItem=str_replace(array("_"," ","/"),"",$gotItem);
if($item==$gotItem){
$rate= $getRates['rate'];
$totRate=$val*$rate;
return $totRate;
}
}
}
The Result Call PHP file
$query = "SELECT * FROM rates ORDER BY item";
$result = $orderdb->query($query)
if (isset($_POST[$itemname]) && !empty($_POST[$itemname])) {
$val=$_POST[$itemname];
total($itemname);
echo $totprate;
} else {
echo "0";
}
I am writing this with the assumption that your SQL is working but are having problems displaying what you want - this may help. The code below saves your $result variable from your query and then passes it into the total function as a second parameter. Previously you were returning $totprate from total but you were not saving it anywhere - it is now saved to the $totprate variable.
Note: I cannot see $orderdb anywhere in your code, I'm assuming you have that in your file and that it is working.
function total($item, $result){
global $val;
global $pid;
global $pitem;
global $prate;
global $totprate;
global $gotitem;
global $getratess;
// global variable for $result removed so it doesn't overwrite variable passed to function
while($getratess=$result->fetch_assoc()){
$gotitem= strtolower(preg_replace('/[^(\x20-\x7F)]*/',"",$getratess['item']));
$gotitem=str_replace(array("_"," ","/"),"",$gotitem);
if ($item==$gotitem) {
$pid=$getratess['id'];
$pitem= $getratess['item'];
$prate= $getratess['rate'];
$totprate=$val*$prate;
return $totprate;
}
}
}
$query = "SELECT * FROM rates ORDER BY item";
$result = $orderdb->query($query);
if (isset($_POST[$itemname]) && !empty($_POST[$itemname])) {
$val=$_POST[$itemname];
$totprate = total($val, $result); // pass itemname as first parameter and result array as second parameter and save it to the $totprate variable
echo $totprate;
} else {
echo "0";
}
Let me know if this helps.
hello i want to create function with returning data, for example when i have the function advert i want to make it every time show what i need, i have the table id, sub_id, name, date, and i want to create the function that i can print every time what i need advert(id), advert(name), i want to make it to show every time what i need exactly and i want to save all my result in array, and every time grab the exactly row that i want
<?php
function advert($data){
$id = $_GET['id'];
$query = mysql_query("SELECT *FROM advertisement WHERE id = $id");
while($row = mysql_fetch_assoc($query)){
$data = array(
'id' => $row['id']
);
}
return $data;
}
echo advert($data['id']);
?>
but my result every time is empty, can you help me please?
There are so many flaws in this short piece of code that the only good advice would be to get some beginners tutorial. But i'll put some effort into explaining a few things. Hopefully it will help.
First step would be the line function advert($data), you are passing a parameter $data to the method. Now later on you are using the same variable $data in the return field. I guess that you attempted to let the function know what variable you wanted to fill, but that is not needed.
If I understand correctly what you are trying to do, I would pass in the $id parameter. Then you can use this function to get the array based on the ID you supplied and it doesnt always have to come from the querystring (although it could).
function advert($id) {
}
Now we have the basics setup, we want to get the information from the database. Your code would work, but it is also vulnerable for SQL injection. Since thats a topic on its own, I suggest you use google to find information on the subject. For now I'll just say that you need to verify user input. In this case you want an ID, which I assume is numeric, so make sure its numeric. I'll also asume you have an integer ID, so that would make.
function advert($id) {
if (!is_int($id))
return "possible SQL injection.";
}
Then I'll make another assumption, and that is that the ID is unique and that you only expect 1 result to be returned. Because there is only one result, we can use the LIMIT option in the query and dont need the while loop.
Also keep in mind that mysql_ functions are deprecated and should no longer be used. Try to switch to mysqli or PDO. But for now, i'll just use your code.
Adding just the ID to the $data array seems useless, but I guess you understand how to add the other columns from the SQL table.
function advert($id) {
if (!is_int($id))
return "possible SQL injection.";
$query = mysql_query("SELECT * FROM advertisement WHERE id = $id LIMIT 1");
$row = mysql_fetch_assoc($query);
$data = array(
'id' => $row['id']
);
return $data;
}
Not to call this method we can use the GET parameter like so. Please be advised that echoing an array will most likely not give you the desired result. I would store the result in a variable and then continue using it.
$ad = advert($_GET['id']);
if (!is_array($ad)) {
echo $ad; //for sql injection message
} else {
print_r($ad) //to show array content
}
Do you want to show the specific column value in the return result , like if you pass as as Id , you want to return only Id column data.
Loop through all the key of the row array and on matching with the incoming Column name you can get the value and break the loop.
Check this link : php & mysql - loop through columns of a single row and passing values into array
You are already passing ID as function argument. Also put space between * and FROM.
So use it as below.
$query = mysql_query("SELECT * FROM advertisement WHERE id = '".$data."'");
OR
function advert($id)
{
$query = mysql_query("SELECT * FROM advertisement WHERE id = '".$id."'");
$data = array();
while($row = mysql_fetch_assoc($query))
{
$data[] = $row;
}
return $data;
}
Do not use mysql_* as that is deprecated instead use PDO or MYSQLI_*
try this:
<?php
function advert($id){
$data= array();
//$id = $_GET['id'];
$query = mysql_query("SELECT *FROM advertisement WHERE id = $id");
while($row = mysql_fetch_assoc($query)){
array_push($data,$row['id']);
}
return $data;
}
var_dump($data);
//echo advert($data['id']);
?>
I'm trying to create a drop down list that populates a <select> with options pulled from a DISTINCT argument. Code looks like this:
function cityData() {
$db =& JFactory::getDBO();
$query = "SELECT DISTINCT MSTCITY FROM " . $db->nameQuote('#__mls') . " ORDER BY MSTCITY;";
$db->setQuery($query);
$tbl = $db->loadObjectList();
echo $tbl;
}
Now, I have two views: one is RAW for an AJAX call and the other is the default view. I figured the simplest way would be to just use the default view and do it in PHP, since the default view wasn't really being used for much anyway. So I added a function:
function dropList($tpl = null){
$model = &$this->getModel();
$array = $model->cityData();
$this->assignRef('array', $array );
parent::display($tpl);
}
And then a call in the page
<?php
$thing = $this->array;
echo $thing;
?>
Nothing is being displayed for the echo $thing;. In the past, when I used PHP to build content instead of AJAX, this worked fine. I don't know if it's using loadObjectList() that's not giving me anything or what. I know the mySQL query works, as it's be tested in the cmd and I get the result I expect.
In order to debug and see the array values you need to use print_r.
In your case edit the default.php file to
<?php
$thing = $this->array;
print_r($thing);
?>
You can also use var_dump
Hi guys I´m new at stackoverflow and also new at Jquery
Well hope I can make myself understandable. Here is what I want: I have made a query to my MySQL db, using a class with PHP
public function User($id) {
$this->connect_db_web($conn);
$sql = mysql_query("SELECT * FROM users WHERE id='".$id."'");
while ($values = mysql_fetch_array($sql)) {
$arr[]=array(
'id'=>$values['idUsers'],
'name'=>$values['name'],
'name2'=>$values['name2'],
'lname'=>$values['lname'],
'lname2'=>$values['lname2'],
'email'=>$values['email'],
'phone'=>$values['phone'],
'address'=>$values['address'],
'bday'=>$values['bday'],
'password'=>$values['password']
);
}
echo '{"user":'.json_encode($arr).'}';
}
Then I have a php code where I call this function
$name = $user->User($id);
I think this works ok (if I´m wrong please help). Now what I´m really trying to do is getting the values from the JSON array into specific divs, example:
$.getJSON("user.php",function(data){
$.each(data.user, function(i,user){
name = user.name;
$(name).appendTo('#getname');
});
});
And inside my HML i Have a <p id="getname"></p>wich is the tag I want the value to be displayed
But no value is displayed, why?, what am I doing wrong?
Thanks for the help I apreciate it
Your JSON is malformed. You are appending a bunch of objects {.1.}{.2.}{.3.}. Instead, try {"users":[{.1.},{.2.},{.3.}]}.
In PHP you'll do something like this (note that I've changed the response type to JSON-P rather than JSON by adding a callback parameter):
public function User($id) {
$users = array();
$this->connect_db_web($conn);
$sql = mysql_query("SELECT * FROM users WHERE id='".$id."'");
while ($values = mysql_fetch_array($sql)) {
$users[] = array(
'id'=>$values['idUsers'],
'name'=>$values['name']
// etc.
);
}
$obj['users'] = $users;
$callback = (empty($_GET["callback"])) ? 'callback' : $_GET["callback"];
echo $callback . '(' . json_encode($obj) . ');';
}
Then you'll be able to do:
$.getJSON("user.php?callback=",function(data){
$.each(data.users, function(i,user){
$('#getname').append(user.name);
});
});
probably safer to do like this:
echo json_encode(array("user" => $arr));
on the other end you would receive an object which, I would suggest iterating like this:
var k;
for (k in data.user){
$("#getname").append($("<span></span>").html(data.user[k].name));
}
Given that you are fetching information for one user only, following I would suggest
$id = (int) $_GET["id"]; // or wherever you get it from.
if ($r = $db->mysql_fetch_assoc()){
$response = array(
"name" => $r["name"];
);
echo json_encode($response);
} else {
echo json_encode(array("error" => "Could not get name for user " . $id));
}
Then, on front-end, all you need to do is:
if (typeof(data.name) != "undefined"){
$("#getname").html(data.name);
} else if (typeof(data.error) != "undefined"){
$("#getname").html(data.error); //or handle otherwise
}
You've misinterpreted your JSON structure. You're appending your DB rows to an array, and embedding that inside an object. If you'd do a console.log(user) inside your .getJSON call, you'd see you'll have to do:
user[0].name
instead. As well, your code assumes that the user ID exists, and returns data regardless of how many, or how few, rows there actually are in the result set. At minimum your JS code code should check users.length to see if there ARE are any rows to begin with. Beyond that, unless you're doing it in another section of code somewhere, that $id value is probably coming from the web page, which means your query is vulnerable to SQL injection attacks.
OK got it,
was a php code error and JSON structre as marc said, here I´m gonna post what finally I had
PHP Class
public function User() {
$users = array();
$this->connect($conn);
$sql = mysql_query("SELECT * FROM users WHERE id='1'");
$values = mysql_fetch_array($sql);
$users[] = array(
'id'=>$values['id'],
'name'=>$values['name'],
'name2'=>$values['name2'],
'lname'=>$values['lname'],
...//rest of values
);
echo json_encode($users);
}
PHP module to get class
include"class.php";
$user = new Users();
$user->User();
Now how did I got the values using JQuery
$.getJSON('user.php', function(data){
$('wherever_you_want_to_point_at').text(data[0].name);
});
Hope it helps someone,
Thanks again guys, very very helpful
Take care you all