Using Chain SELECT works great from SELECT to SELECT, I'm trying to do SELECT to INPUT.
My mainpage.php
<label>Manufacturer</label>
<select>My Select statement is here</select>
<label>Model</label>
<select name="modelname">My Select statement is fed from the select above</select>
<label>Rating</label>
<input name="rating"></input>
This is the jQuery I have in the <head> section on the mainpage.php
<script>
$(document).ready(function(){
$("select#modelname").change(function(){
var id = $("select#modelname option:selected").attr('value');
$.post("assets/configs/getdata.php", {id:id}, function(data){
$("input[name='rating']").html(data);
console.log(data);
});
});
});
</script>
and finally the getdata.php
<?php
include "db.php";
$modelid = $_POST[id];
$sql = "SELECT EfficiencyRating FROM AllModels WHERE ModelID = '$modelid' ";
$res = odbc_exec($cnn, $sql);
while ($row = odbc_fetch_array($res)) {
$row_array[] = $row['EfficiencyRating'];
array_push($return_arr,$row_array);
}
echo json_encode($return_arr);
?>
Using the console log when this message is returned, how can I fix this?
HP Warning: array_push() expects parameter 1 to be array, null given in assets\configs\getdata.php on line 12
Try with this.
$res = odbc_exec($cnn, $sql);
$return_arr = array();
while ($row = odbc_fetch_array($res)) {
$return_arr[] = $row['EfficiencyRating'];
}
echo json_encode($return_arr);
JS Part
// Slightly modify the Request
$.post("assets/configs/getdata.php", {id:id}, function(data){
// JSON Object
console.log(data);
$("input[name='rating']").val(data);
}, 'json');
You need to declare $return_arr before the while statement. Also, I personally feel what you are doing is just not right. The proper way would be this...
$res = odbc_exec($cnn, $sql);
$return_arr = array(); //<----------- Here
while ($row = odbc_fetch_array($res)) {
array_push($return_arr,$row['EfficiencyRating']);
}
echo json_encode($return_arr);
Related
I would like to update the data in the frontend when it is changed in the database. The code I'm using is given below:
<script src="https://code.jquery.com/jquery-3.5.0.js"></script>
<div id="test">
<?php
include('conn.php');
$query = "SELECT name FROM user_details WHERE user_id = 1;";
if(mysqli_fetch_assoc(mysqli_query($conn, $query))["name"] == "MyName")
echo 'Hi <b>MyName!</b>';
else
echo 'You are not <b>MyName</b>.';
?>
</div>
<script>
setInterval(function(){
$.get("/test.php", function(data){
let $data = $(data);
$("#test").append($data.find("#test > *"));
});
}, 1000);
</script>
However, when the data is updated, it does not get updated in the frontend unless refreshed. When I use jQuery's load() function, it works perfectly. Why does this not work?
As I suggested in the comment, if you create a stand alone PHP Script, it might be like:
getUserName.php
<?php
$id = (int)$_GET['id'];
include('conn.php');
$query = "SELECT name FROM user_details WHERE user_id = $id;";
$myName = "";
if ($result = mysqli_query($conn, $query)) {
while ($row = mysqli_fetch_assoc($result)) {
$myName = $row;
}
}
mysqli_free_result($result);
mysqli_close($conn);
header('Content-Type: application/json');
echo json_encode($myName);
?>
This is a very basic example and I would strongly advise you switch to using prepared statements to avoid the risk of SQL Injection.
In your HTML you can now do:
<script>
setInterval(function(){
$.getJSON("/getUserName.php", { id: 1 }, function(data){
$("#test").append(data.name);
});
}, 1000);
</script>
This will ping the script every second and you will have a list of names appearing.
I'm new in Ajax and JSON notation, so I'm trying to get data from differents tables of a Database, data like country names, state names, departament name, job position etc. and I've seen examples how through JSON can get data but just from a single table, can you give me a little help how can I do it with more than one table and keep it in an array.
<?php
$host = "localhost";
$user = "usuer";
$pass = "password";
$databaseName = "jsonExample";
$tableName = "variables";
$con = mysql_connect($host,$user,$pass);
$dbs = mysql_select_db($databaseName, $con);
$result = mysql_query("SELECT * FROM $tableName"); //query
//$array = mysql_fetch_row($result); //fetch result
if(mysql_num_rows($result) <= 0){
}else{
while($obj = mysql_fetch_row($result)){
$array[] = $obj;
}
}
echo json_encode($array);
?>
Html file:
<html>
<head>
<script language="javascript" type="text/javascript" src="jquery.js"></script>
</head>
<body>-->
<h2> Client example </h2>
<h3>Output: </h3>
<div id="output">this element will be accessed by jquery and this text will be replaced</div>
<script id="source" language="javascript" type="text/javascript">
$(function ()
{
$.ajax({
url: 'api.php', //the script to call to get data
data: "", //you can insert url argumnets here to pass to api.php for example "id=5&parent=6"
dataType: 'json', //data format
success: function(data) //on recieve of reply
{
var id = data[0]; //get id
var vname = data[1]; //get name
$('#output').html("<b>id: </b>"+id+"<b> name: </b>"+vname); //Set output element html
//recommend reading up on jquery selectors they are awesome http://api.jquery.com/category/selectors/
}
});
});
</script>
</body>
</html>
If you want to have the results from multiple queries in one array you can add each result to a key. F.i. if you querying table table1 to tablen ...
// define the array that will contain all result sets
$array = [];
// create an array for the result set coming from table 1
$array['table1']= [];
$result = mysql_query("SELECT * FROM table1");
if(mysql_num_rows($result) <= 0){
}else{
while($obj = mysql_fetch_row($result)){
$array['table1'][] = $obj;
}
}
// create an array for the result set coming from table 2
$array['table2']= [];
$result = mysql_query("SELECT * FROM table2");
if(mysql_num_rows($result) <= 0){
}else{
while($obj = mysql_fetch_row($result)){
$array['table2'][] = $obj;
}
}
::
::
// create an array for the result set coming from table n
$array['tablen']= [];
$result = mysql_query("SELECT * FROM tablen");
if(mysql_num_rows($result) <= 0){
}else{
while($obj = mysql_fetch_row($result)){
$array['tablen'][] = $obj;
}
}
// return the results formatted as json
return json_encode($array);
In javascript you can access the results for table1 with data->table1.
Tip
Use mysqli instead of mysql. It is the improved version of mysql. Check the answers for this question for some background.
I have a jquery save script like :
naam = prompt('Give a name for your file.');
if(naam != null)
{
var div_contents = $("#print").html();
$.post("save.php", { 'contents': div_contents,'naam':naam });
alert('Your file is save as : '+ naam);
window.location.replace("index.php?id=latest");
}
else
{
alert('Not saved');
}
I save a div in save.php which creates an new id in the database
What I want to achive is were
window.location.replace("index.php?id=latest");
id=latest must become (id=id from last saved file).
I tried
$q = "select MAX(id) from Moodboards";
$result = mysql_query($q);
$data = mysql_fetch_array($result);
$MBId = $data[0];
window.location.replace("index.php?id="+MBId);
and
var MBID =
<?php
$q = "select MAX(id) from Moodboards";
$result = mysql_query($q);
$data = mysql_fetch_array($result);
$MBId = $data[0];
echo $MBId ?>
window.location.replace("index.php?id="+MBId);
They both failed.
How can I run the query in the if(naam !=null) statement?
At first place you must fix your jQuery POST... You don't use POST respond which is wrong.. You should wait for it and then continue with other actions
naam = prompt('Give a name for your file.');
if(naam != null)
{
var div_contents = $("#print").html();
$.post("save.php", { 'contents': div_contents,'naam':naam }, function(responde){
if(responde.id)
window.location.replace("http://yoururl.com/index.php?id="+responde.id);
else
alert("No responde...");
}, "json");
}
else
{
alert('Not saved');
}
For better results I suggest you to use JSON data in that post/respond..
At your PHP code you have to set:
<?php
$q = "select MAX(id) from Moodboards";
$result = mysql_query($q);
$data = mysql_fetch_array($result);
$MBId = $data[0];
echo json_encode(array('id'=>$MBId));
exit();
?>
P.S. For window.location.replace please set your FULL url: "http://localhost/index.php?id=" OR atleast put slash at start of it "/index.php?id="
Solution
if(naam != null)
{
var div_contents = $("#print").html();
$.post("save.php", { 'contents': div_contents,'naam':naam });
alert('Uw moodboard is opgeslagen als '+ naam);
window.location.replace("index.php?id=<?php $q = "select MAX(id) from Moodboards";
$result = mysql_query($q);
$data = mysql_fetch_array($result);
$MBId = ($data[0] + 1); echo "$MBId";?>");
}
This Works for me , i didnt need to make a jquery var i could echo the variable in php.
And i had to add 1 cause the sql query is loaded when the page is loaded.
So the file isn't saved yet when i get the highest id.
I am trying to use getJSON to load the ids that are stored into a database, i am passing the values from PHP to JavaScript by echoing it out. I am trying to alert the value that i passed through but it didn't work
The PHP and JavaScript are on the same page. Any guidance would be appreciated
I printed the JSON file and it is in this format
Array
(
[0] => 213
[1] => 214
[2] => 215
)
PHP
$result = mysql_query("SELECT * FROM address");
$arra = array();
while ($row = mysql_fetch_array($result)){
$arra[] = $row['id'];
}
Javascript
<script>
var test= <?php echo json_encode($arra); ?>
var url = test;
$.getJSON(url, function(data) {
alert(url);
})
</script>
Make new php file called "Data.php" and make it return your JSON data.
<?php
$con = mysql_connect('localhost', 'root', '');
mysql_select_db('test');
$result = mysql_query("SELECT * FROM address ORDER BY id DESC");
$arra = array();
while ($row = mysql_fetch_array($result)){
$arra[] = $row['id'];
}
echo json_encode($arra);
?>
Then where you call your $.getJSON, Do:
$.getJSON(data.php, function(data) {
alert(data);
})
url is json object in your case. It should be url of a page! something like this should work:
$.getJSON("<?=$_SERVER["PHP_SELF"]?>", test, function(data) {
alert(data);
})
I have to make a JSON ajax request and have an array in return.
i found this code on multiple other questions (edited to my problem):
var hej[];
function ajax_search(){
$.getJSON('test.php',function(response){
var data = response.get_response;
for (i in data){
hej.push([i,data[i]]);
}
alert(hej[0]);
}
In the php part, i have a database from which i need the data i have have 5 columns pr. row. I only have one row, and i need all five columns in the array.
$sql = "SELECT * FROM 'table'
LIMIT 1 OFFSET $i"; //random offset
$result = mysql_query($sql);
$storeArray = Array();
while($row = mysql_fetch_array($result))
{
$storeArray[0]=$row['column1'];
$storeArray[1]=$row['column2'];
$storeArray[2]=$row['column3'];
$storeArray[3]=$row['column4'];
$storeArray[4]=$row['column5'];
}
$json = json_encode($storeArray);
echo $json;
I know im doing something wrong, and i am new to programming. I need to be able to access all column values in my javascript after the call and use them in other functions, but i cant get it to return.
I would really appriciate help, with both the javascript and the php if there are errors in either.
It should be:
function ajax_search(){
$.getJSON('test2.php',function(response){
// response.column1
// response.column2
// response.column3
// for array push:
var hej = [];
$.each(response, function(key, val) {
hej.push(val);
});
alert(hej[0]); // it will alert column1
});
}
and MySQL:
$sql = "SELECT `column1`, `column2`, `column3`, `column4`, `column5` FROM `table` LIMIT 1 OFFSET $i"; //random offset
$result = mysql_query($sql);
$row = mysql_fetch_assoc($result);
$json = json_encode($row);
echo $json;
Try this:
Javascript Code:
var hej[];
function ajax_search(){
$.getJSON('test.php',function(response){
$.each(response.results, function(key, val) {
hej.push(val);
});
alert(hej[0]);
});
}
PHP Code:
<?php
$sql = "SELECT * FROM 'table' LIMIT 1 OFFSET $i"; //random offset
$result = mysql_query($sql);
$storeArray = array();
while($row = mysql_fetch_array($result)) {
$storeArray[0]= $row['column1'];
$storeArray[1]=$row['column2'];
$storeArray[2]=$row['column3'];
$storeArray[3]=$row['column4'];
$storeArray[4]=$row['column5'];
}
$json = json_encode( array('results' => $storeArray ) );
echo $json;
?>
Hope this helps.
Why use response.get_response?
var hej[];
function ajax_search(){
$.getJSON('test.php',function(data){
$.each(data, function (i,item) {
hej.push([i,item]);
}
alert(hej[0]);
}