I'm newbie with AJAX, and I found the code solution for my problem in this page, but I can't make it work.
I need that the input #precio changes when I select an option in select #producto.
I have this code:
<form action="actualizarprod_pr.php" method="post">
<input type="text" name="producto" value="" placeholder="Producto" list="productolist" />
<datalist id="productolist">
<select id="productos" style="display: none;">
<?php
$query = "SELECT * FROM lista_precios";
$query_ej = mysqli_query($conexion, $query);
while($registro = mysqli_fetch_array($query_ej)){
echo "<option value='" . $registro['producto'] .
"'</option>";
}
?>
</select>
</datalist>
<input type="text" id="precio" name="precio" placeholder="Precio">
<input type="submit" name="Actualizar" value="Actualizar">
<input type="submit" name="Borrar" value ="Borrar">
</form>
Script:
<script type="text/javascript">
$(document).ready(function(){
$("#productos").change(function(){
var producto=$(this).val();
$.ajax({
type:"post",
url:"cargaselect.php",
data:"producto="+producto,
success:function(data){
$("#precio").val(data);
}
});
});
});
</script>
cargaselect.php:
<?php
session_start();
include("conexion.php");
if(isset($_SESSION["id_usuario"])){
$producto=$_POST["producto"];
$result = mysqli_query($conexion,
"select * FROM lista_precios WHERE producto =". $producto ." ");
echo $result[precio];
} else {
header("location: login.php");
}
?>
This line is not formatted correctly
echo $result[precio];
should be
echo $result['precio'];
What does $result['precio'] return?
If its an array of results you might need to use
$result[0]['precio'] perhaps instead
Related
I have a small problem.
<select name="level_id"> is not posting.
So i get an error like : Undefined index : level_id
Exactly what can i do?
$sql1 = 'SELECT T_ABILITY.PK AS AB_PK,T_ABILITY.ABILITY_NAME AS AN,T_ABILITY_LEVEL.PK AS LE_PK,T_ABILITY_LEVEL.LEVEL_NAME AS LN
FROM T_USER_ABILITY_REL,T_ABILITY,T_ABILITY_LEVEL WHERE
T_USER_ABILITY_REL.ABILITY_FK = T_ABILITY.PK AND
T_USER_ABILITY_REL.ABILITY_LEVEL_FK = T_ABILITY_LEVEL.PK AND
T_USER_ABILITY_REL.USER_FK = '.$user_id.'
ORDER BY AN';
$stmt1 = oci_parse($conn, $sql1);
$r1 = oci_execute($stmt1);
while ($row1 = oci_fetch_array($stmt1, OCI_RETURN_NULLS + OCI_ASSOC)) {
echo '<form method="post">';
echo '<tr>';
echo '<td>'.$row1["AN"].'</td>';
echo '<input type="hidden" name="ability_id" value="'.$row1["AB_PK"].'"/>';
echo '<td class="select-level">';
$sql2 = 'SELECT PK,LEVEL_NAME FROM T_ABILITY_LEVEL ORDER BY LEVEL_ORDER';
$stmt2 = oci_parse($conn, $sql2);
$r2 = oci_execute($stmt2);
echo '<select name="level_id" class="form-control selectpicker" data-container="body" data-live-search="true" data-size="5" title="Seviye Seçiniz">';
while ($row2 = oci_fetch_array($stmt2, OCI_RETURN_NULLS + OCI_ASSOC)) {
echo '<option '.($row2["PK"] == $row1["LE_PK"] ? 'selected="selected"' : "").' value="'.$row2["PK"].'">'.$row2["LEVEL_NAME"].'</option>';
}
echo '</select>';
echo '<button type="submit" name="update-user-ability" class="btn btn-success">Güncelle</button>';
echo '<button type="submit" name="delete-user-ability" class="btn btn-danger">Sil</button>';
echo '</td>';
echo '</tr>';
echo '</form>';
}
submit part below
if (isset($_POST["update-user-ability"])) {
$user_id = $_GET["user_id"];
$ability_id = $_POST["ability_id"];
$level_id = $_POST['level_id'];
This is because your select is empty (that is, the oci_fetch_array($stmt2, OCI_RETURN_NULLS + OCI_ASSOC returns no rows.)
A select, when no option are present, will not return a value to the receiving php script.
Add a default option before the loop to be sure something is passed even if your sql query returns nothing.
this script will return an empty post:
<html>
<body>
<form method="post">
<select name="select"></select>
<input type="submit">
</form>
</body>
</html>
<?php
var_dump($_POST); // array(0) { }
while this one will have the value 12 defined:
<html>
<body>
<form method="post">
<select name="select">
<option value="12">12</option>
</select>
<input type="submit">
</form>
</body>
</html>
<?php
var_dump($_POST); // array(1) { ["select"]=> string(2) "12" }
Could be that the value of level_id isn't send along, because it is a <select>. Not sure, but could be that only <input /> values are properly posted.
You could use a <input type="hidden" name="level_id_val" /> and update its value with JavaScript every time the value of <select name="level_id"> is changed (using .onchange).
Then, in PHP, use that field instead of the select: $level_id = $_POST['level_id_val'];
I am new to Angular and am struggling to add a comment to a specific post that is being displayed from a database.
Under the post I have a hidden comment area that shows when pressing the button.
I then want to be able to add the comment to the post the comment area is attached too.
It seems like it is not getting the id but I can't find how to solve it.
It is doing the show/hide, and I am getting the "data inserted" in the console log so I'm guessing the problem is with the PHP?
Angular code
app.controller('meetings', function($scope, $http) {
$scope.meetings_insertdata = function() {
$http.post ("meetings_insert.php",{'meetings_commentdate':$scope.meetings_commentdate,'meetings_comment':$scope.meetings_comment})
window.location.reload();
console.log("data inserted");
};
});
app.controller('show_hide', function ($scope) {
$scope.Hidden = true;
$scope.ShowHide = function () {
$scope.Hidden = $scope.Hidden ? false : true;
}
});
meetings_insert.php
<?php
$data = json_decode(file_get_contents("php://input"));
$meetings_comment = $data->meetings_comment;
$meetings_commentdate = date('Y-m-d H:i:s');
$meetings_entry = $_GET['id'];
mysql_connect("localhost","user","password");
mysql_select_db("mdb_rh316");
mysql_query("UPDATE project_meetings SET (meetings_comment, meetings_commentdate) = ('$meetings_comment','$meetings_commentdate') WHERE meetings_entry = '$meetings_entry'");
?>
HTML code
<div class="entrybox" ng-controller="meetings">
<?php
mysql_connect("localhost","user","password");
mysql_select_db("mdb_rh316");
$result = mysql_query("SELECT * FROM project_meetings ORDER BY meetings_date DESC");
while ($row = mysql_fetch_array($result))
{
echo "<table>";
echo "<tr><td>" ?>Added: <? echo $row['meetings_date'] . $row['meetings_entry'] . "<br/>" . $row['meetings_content']."</td></tr>";
echo "<tr><td>" ?><br/><span class="emp">Comment: <?
echo $row['meetings_commentdate']."<br/>" . $row['meetings_comment']."</td></tr>";
echo "<tr><td>"?></span>
<div ng-controller="show_hide">
<input type="button" class="previous_add" value="Add comment" ng-click="ShowHide()" />
<br />
<br />
<div ng-hide = "Hidden">
<form method="post" action="meetings_insert.php?id=<? echo $row['$meetings_entry']?>">
<textarea class="textarea" placeholder="Add your comment here.." type="text" ng-model="meetings_comment" name="meetings_comment"></textarea><br/>
<input type="button" class="button" value= "Add" ng-click="meetings_insertdata()"/>
</form>
</div>
</div>
</td></tr><br/><?;
}
echo "</table>";
?>
</div>
Pass 'id' into the meetings_insertdata function:
<form method="post" action="">
<textarea class="textarea" placeholder="Add your comment here.." type="text" ng-model="meetings_comment" name="meetings_comment"></textarea>
<br/>
<input type="button" class="button" value="Add" ng-click="meetings_insertdata(<? echo $row['$meetings_entry']?>)" />
</form>
Receive it in the AngularJS function below:
app.controller('meetings', function($scope, $http) {
$scope.meetings_insertdata = function(id) {
$http.post("meetings_insert.php", {
'meetings_commentdate': $scope.meetings_commentdate,
'meetings_comment': $scope.meetings_comment,
'meetings_event': id
})
window.location.reload();
console.log("data inserted");
};
});
Then, pick up 'id' from the posted value ($data->meetings_event)
$data = json_decode(file_get_contents("php://input"));
$meetings_comment = $data->meetings_comment;
$meetings_commentdate = date('Y-m-d H:i:s');
$meetings_entry = $data->meetings_event;
Hi i want to update my sql table field without page refresh in php how can i achieve i tried but my code is not working i do not know where i am wrong
index.php
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN"
"http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<script src="http://code.jquery.com/jquery-1.9.1.js"></script>
<script>
$(document).ready(function(){
$('#myForm').on('submit',function(e) {
$.ajax({
url:'assignlead.php',
data:$(this).serialize(),
type:'POST',
success:function(data){
console.log(data);
$("#success").show().fadeOut(5000);
},
error:function(data){
$("#error").show().fadeOut(5000);
}
});
e.preventDefault();
});
});
</script>
</head>
<body>
<?php
include('conn.php');
$per_page = 3;
if($_GET)
{
$page=$_GET['page'];
}
$start = ($page-1)*$per_page;
$select_table = "select * from clientreg order by id limit $start,$per_page";
$variable = mysql_query($select_table);
?>
<form class="form2" action="" method="POST" name="myForm" id="myForm">
<div style="width:100%;">
<?php
$i=1;
while($row = mysql_fetch_array($variable))
{
?>
<input type="checkbox" name="users[]" value="<?php echo $row["id"]; ?>" >
<?php
}
?>
<div class="buttons"> <span id="error" style="display:none; color:#F00">Some Error!Please Fill form Properly </span> <span id="success" style="display:none; color:#0C0">All the records are submitted!</span>
<input class="greybutton" type="submit" value="Send" />
</div>
<?php
$sql = mysql_query("SELECT *FROM login where role=1");
while ($row = mysql_fetch_array($sql)){
?>
<input type="checkbox" name="eid[]" value="<?php echo $row["eid"]; ?>" ><?php echo $row["username"]; ?>
<?php
}
?>
</div>
</form>
</div>
</body>
</html>
assignlead.php
<?php
$conn = mysql_connect("localhost","root","root");
mysql_select_db("helixcrm",$conn);
if(isset($_POST["submit"]) && $_POST["submit"]!="") {
$usersCount = count($_POST["id"]);
for($i=0;$i<$usersCount;$i++) {
mysql_query("UPDATE clientreg set eid='" . $_POST["eid"][$i] . "' WHERE id='" . $_POST["id"][$i] . "'");
}
}
?>
<?php
$rowCount = count($_POST["users"]);
for($i=0;$i<$rowCount;$i++) {
$result = mysql_query("SELECT * FROM clientreg WHERE Id='" . $_POST["users"][$i] . "'");
$row[$i]= mysql_fetch_array($result);
$id=$row[$i]['id'];
?>
<input type="hidden" name="id[]" class="txtField" value="<?php echo $row[$i]['id']; ?>"></td>
<?php
$rowCoun = count($_POST["eid"]);
for($j=0;$j<$rowCoun;$j++) {
$result = mysql_query("SELECT * FROM login WHERE eid='" . $_POST["eid"][$j] . "'");
$row[$j]= mysql_fetch_array($result);
$eid=$row[$j]['eid'];
?>
<input type="hidden" name="eid[]" class="txtField" value="<?php echo $row[$j]['eid']; ?>">
<?php
}
}
?>
i tried a lot but i am not able to get my output
How can i achieve my output
Thanks in advance
Update your javascript like this :
$(document).ready(function(){
$('#myForm').submit(function(){
$.ajax({
url : 'assignlead.php',
data : $(this).serialize(),
type : 'POST',
success : function(data){
console.log(data);
$("#success").show().fadeOut(5000);
},
error:function(data){
$("#error").show().fadeOut(5000);
}
});
// !important for ajax form submit
return false;
});
});
I'm creating a page which will allow an admin to select a user from a drop down list, which populates from a database. When the person is selected, the info associated with that person will then be viewed on the page. I already have a select statement which selects all the info and the drop down menu is populating correctly. However, I'm unsure on how to get that selected user's info to display on the page once selected. Would I need to do an entirely different select statement and query which checks which customer was selected? Or is there another way?
customer.php
<div id="view_form" class="view">
<form method="post" action="<?php echo htmlspecialchars($_SERVER["PHP_SELF"]);?>">
<fieldset>
<label for="viewCustomer">Select Customer</label>
<?php
echo "<select name='selectCust' id='selectCust'>";
echo "<option></option>";
while($row = mysqli_fetch_assoc($custResult)){
$name = "{$row['fName']} {$row['lName']}";
echo "<option>$name</option>";
}
echo "</select>";
?>
</fieldset>
</form>
</div>
viewUser.php
if(isset($search)){
$select = "SELECT * FROM $cust WHERE acctNum='{$search}'";
$result = mysqli_query($db, $select);
if(mysqli_num_rows($result) > 0){
if($row = mysqli_fetch_assoc($result)){
$acct = "{$row['acctNum']}";
echo $acct;
}
}
}
script.js
$(document).ready(function(){
function searchAjax(){
var search = $('#selectCust').val();
$.post('includes/viewUser.php', {searchUsers: search}, function(data){
$('#view_form').append(data);
})
}
$('#selectCust').on('change', function(ev){
ev.preventDefault();
searchAjax();
})
})
Search.php
<script type="text/javascript "src="//ajax.googleapis.com/ajax/libs/jquery/1.8.0/jquery.min.js"></script>
<script type='text/javascript'>
$(document).ready(function(){
$(".dropdown-users").on("change",function(event){
event.preventDefault();
search_ajax_way();
});
});
function search_ajax_way(){
var search_this=$("dropdown-users").val();
$.post("Ajaxsearch.php", {searchusers : search_this}, function(data){
$(".results").html(data);
})
}
</script>
<div id="view_form" class="view">
<form method="post">
<fieldset>
<label for="viewCustomer">Select Customer</label>
<?php
echo "<select class="dropdown-users">";
echo "<option></option>";
while($row = mysqli_fetch_assoc($custResult)){
$name = "{$row['fName']} {$row['lName']}";
$acct = $row['acctNum'];
echo "<option value="$acct">$name ($acct)</option>";
}
echo "</select>";
?>
</fieldset>
</form>
</div>
<label>Enter</label>
<input type="text" name="search_query" id="search_query" placeholder="What You Are Looking For?" size="50"/>
<input type="<span id="IL_AD1" class="IL_AD">submit</span>" <span id="IL_AD6" class="IL_AD">value</span>="Search" id="button_find" />
<div class="results"></div>
//********************************************************************************************
********************************************************************************************//
Ajaxsearch.php
<?php
$con = mysqli_connect("localhost","my_user","my_password","my_db"); // Enter your information here
$term = $_POST['searchusers']
$term = mysqli_real_escape_string($con, $term);
if($term == "")
echo "Enter Something to search";
else {
$query = mysqli_query($con, "select * from USERDATEBASEHERE where ID = '{$term}' ");
$string = '';
if (mysqli_num_rows($query) > 0) {
if (($row = mysqli_fetch_assoc($query)) !== false) {
$string = "{$row['ID']}";
}
} else {
$string = "This Person does not exist";
}
echo $string;
}
?>
<div id="view_form" class="view">
<form method="post" action="<?php echo htmlspecialchars($_SERVER["PHP_SELF"]);?>">
<fieldset>
<label for="viewCustomer">Select Customer</label>
<?php
echo "<select name=\"somename\" onchange=\"this.form.submit();\">";
echo "<option value=\"\">Select User</option>";
while($row = mysqli_fetch_assoc($custResult)){
$name = "{$row['fName']} {$row['lName']}";
$acct = $row['acctNum'];
echo '<option value="'.$acct.'">$name ($acct)</option>';
}
echo "</select>";
?>
</fieldset>
</form>
</div>
The options must have some refering value, through which you can retrieve the details of selected user, whenever the value of option is not initiated then the default value of the option will be option's label.
I am trying to print the dropdown selected item. I have well displayed the dropdwon list menu. but when i select an option it doesn't print the option. i have tried in many ways. But not yet got! Please help me, this is my following code.
<form name="choose" method="get" action="<?php echo $_SERVER['PHP_SELF']; ?>">
<?php
$query="SELECT id_cat,name FROM `fs01_metier_cat` ORDER BY `fs01_metier_cat`.`id_cat`";
$result = mysql_query($query);
?>
<?php
echo "<select name=category></option>";
while($nt=mysql_fetch_array($result)) {
echo "<option value='".$nt['name']."'>".$nt['name']."</option>";
}
echo "</select>";
?>
<input type="submit" name="submit" value="save category" />
</form>
<?php
if($_GET){
echo 'The year selected is'.$_GET['category'];
}
?>
You have issues in your code, try this one instead :
<form name="choose" method="get" action="<?php echo $_SERVER['PHP_SELF']; ?>">
<?php
$query="SELECT id_cat,name FROM `fs01_metier_cat` ORDER BY `fs01_metier_cat`.`id_cat`";
$result = mysql_query($query);
?>
<select name=category>
<?php
while($nt=mysql_fetch_array($result)) {
echo "<option value='".$nt['name']."'>".$nt['name']."</option>";
}
?>
</select>
<input type="submit" name="submit" value="save category" />
</form>
<?php
if($_GET){
echo 'The year selected is'.$_GET['category'];
}
?>
$_GET['category']
should be
$_POST['category']
Example for javascript:
<html>
<head>
<script type="text/javascript">
window.onload = function() {
var eSelect = document.getElementById('cat');
eSelect.onchange = function() {
document.getElementById("displaytext").innerHTML = "Selected Value: "+this.value;
document.getElementById("displaytext").style.display= 'block';
}
}
</script>
</head>
<body>
<select id="cat" name="cat">
<option value="x">X</option>
<option value="y">Y</option>
<option value="other">Other</option>
</select>
<div id="displaytext" style="display: none;" ></div>
</body>
</html>