I try to fill text boxes from my database. I am using a onChange event. I can see that the code is running but I get a wrong response.
I am using the following code:
index.php
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "database";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT * FROM cus";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
echo "<select class='form-control select2' name='cus_id' onChange='getCus(this.value)' style='width: 100%;'>";
echo "<option selected disabled hidden value=''></option>";
// output data of each row
while($row = $result->fetch_assoc()) {
echo "<option value='" . $row["cus_id"]. "'>" . $row["cus_id"]. " | " . $row["cus_name"]. "</option>";
}
echo "</select>";
} else {
echo "0 results";
}
$conn->close();
?>
<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script>
<script>
function getCus() {
// getting the selected id in combo
var selectedItem = jQuery('#cus_id option:selected').val();
// Do an Ajax request to retrieve the product price
jQuery.ajax({
url: 'get5.php',
method: 'POST',
data: {'cus_id': selectedItem},
success: function(response){
// and put the price in text field
jQuery('#cus_id').val(response);
},
error: function (request, status, error) {
alert(request.responseText);
},
});
}
</script>
get5.php
<?php
// Turn off all error reporting
error_reporting(0);
?>
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "database";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname) ;
// Check connection
if ($conn->connect_error)
{
die('Connection failed: ' . $conn->connect_error) ;
}
else
{
$cus_name = isset($_POST['cus_name'])?$_POST['cus_name']:'';
$cus_ad = isset($_POST['cus_ad'])?$_POST['cus_ad']:'';
$cus_pos = isset($_POST['cus_pos'])?$_POST['cus_pos']:'';
$cus_pla = isset($_POST['cus_pla'])?$_POST['cus_pla']:'';
$cus_cou = isset($_POST['cus_cou'])?$_POST['cus_cou']:'';
$query = 'SELECT * FROM cus WHERE cus_id=' . $cus_id . ' ';
$res = mysqli_query($conn, $query) ;
if (mysqli_num_rows($res) > 0)
{
$result = mysqli_fetch_assoc($res) ;
echo $result['cus_name'];
echo $result['cus_ad'];
echo $result['cus_pos'];
echo $result['cus_pla'];
echo $result['cus_cou'];
}else{
$result = mysqli_fetch_assoc($res) ;
echo "0 res";
}
}
?>
I cant see whats wrong with my code. When I run this I get the response "0 res". Can anyone see what is wrong?
You have to this line of code
<?php
// Turn off all error reporting
error_reporting(0);
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "database";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname) ;
// Check connection
if ($conn->connect_error) {
die('Connection failed: ' . $conn->connect_error) ;
}else {
$cus_id = isset($_POST['cus_id'])?$_POST['cus_id']:'';
$query = 'SELECT * FROM cus WHERE cus_id="' . mysqli_real_escape_string($conn, $cus_id) . '"';
$res = mysqli_query($conn, $query) ;
if (mysqli_num_rows($res) > 0) {
$result = mysqli_fetch_assoc($res) ;
echo $result['cus_name'];
echo $result['cus_ad'];
echo $result['cus_pos'];
echo $result['cus_pla'];
echo $result['cus_cou'];
}else{
$result = mysqli_fetch_assoc($res) ;
echo "0 res";
}
} ?>
You should always escapes special characters in a string for use in an SQL statement.
Please check your cus_id data type, i guess you didn't set it in integer or another numeric type.
If i am wrong, you should check your row in database, it must be empty in that cus_id.
Related
please can someone help me with this code? It shows all the users’ info but i need it to show only the info of the logged user.
<?php
$servername = "localhost";
$username = "username";
$password = "password";
$dbname = "username";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT AEG FROM users";
$result = $conn->query($sql);
if ($result->num_rows > 3) {
// output data of each row
while($row = $result->fetch_assoc()) {
echo "id: " . $row["AEG"]. "<br>";
}
} else {
echo "0 results";
}
$conn->close();
?>
Upon login, store the email/username/id in a $_SESSION variable;
$_SESSION['email'] = $email; // in this example I used email
Then on your file, you can access session variables using $_SESSION['variable'] and use it on your sql statement;
My modifications are the ones with comments.
<?php
$servername = "localhost";
$username = "username";
$password = "password";
$dbname = "username";
/*Store session data in a variable*/
$email = $_SESSION['email'];
/**********************************/
$conn = new mysqli($servername, $username, $password, $dbname);
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
/*Add where clause to your sql statement*/
$sql = "SELECT AEG FROM users WHERE email ='".$email."'";
/****************************************/
$result = $conn->query($sql);
if ($result->num_rows > 3) {
while($row = $result->fetch_assoc()) {
echo "id: " . $row["AEG"]. "<br>";
}
} else {
echo "0 results";
}
$conn->close();
?>
Try this one if you are having 4 user then it is showing only one
if($result->num_rows > 0)
{
while($row = $result->fetch_assoc())
{
echo "id: " . $row["AEG"]. "<br>";
}
}
else
{
echo "0 results";
}
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "job_database";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error)
{
die("Connection failed: " . $conn->connect_error);
}
if(isset($_POST['Get']))
{
$jtitle = $_POST['jobtitle'];
$location = $_POST['location'];
$category = $_POST['categories'];
//Query specified database for value
$sql = "SELECT * FROM addjob where jtitle ='$jtitle' ∧ location ='$location' ∧ category ='$category' " ;
$result = $conn->query($sql);
}
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
echo "jtitle: " . $row["jtitle"]. "location:" . $row["location"]. "category" . $row["category"]."<br>";
}
} else {
echo "0 results";
}
$conn->close();
?>
Change your sql query as follows,
$sql = "SELECT * FROM addjob WHERE jtitle =".$jtitle." AND location =".$location." AND category =".$category." ;
since ∧ is wrong, what it does is, it ends the SQL query because there is ; in the end. So change ∧ into and and check.
And read more about form validation and SQL injection from these links.
Good luck! =)
What I wan't to do is create buttons that are automatically generated from the database. So when I add a new record in the database the button is created Is this possible with a loop? So yes how do I create the button.
This is what I have so far:
<?php
$servername = "localhost";
$username = "root";
$password = "Iamthebest1009";
$dbname = "dktp";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT * FROM theme";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
echo "". $row["theme_name"]. "<br>";
}
} else {
echo "no results";
}
$conn->close();
?>
Yes it is possible. you need to echo html
<?php
$servername = "localhost";
$username = "root";
$password = "Iamthebest1009";
$dbname = "dktp";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT * FROM theme";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
$your_url ="https://www.google.com";
echo "". $row["theme_name"]. "<br>";
echo '<input type="button" name="' . $row["theme_name"]. '" value="'. $row["theme_name"].'">';
}
} else {
echo "no results";
}
$conn->close();
?>
This is my query, and it is working fine, but how do I have the results returned in an HTML TABLE. for example I want the first 5 results on row 1 the next 5 on row 2 and so on. Any help would be great. Thank you!
<?php
$servername = "XXX";
$username = "XXX";
$password = "XXX";
$dbname = "XXX";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT * FROM manufacturer LIMIT 10 OFFSET 0";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
echo '<td><a href="index.php?route=product/manufacturer/info&
manufacturer_id='. $row["manufacturer_id"] .'"><img src="image/' .
$row["image"] .'"></a></td>';
}
} else {
echo "0 results";
}
$conn->close();
<?php
$servername = "XXX";
$username = "XXX";
$password = "XXX";
$dbname = "XXX";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn- >connect_error);
}
$sql = "SELECT * FROM manufacturer LIMIT 10 OFFSET 0";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
$i=0;
while($row = $result->fetch_assoc()) {
If($i==5){
echo "</tr>";
echo "<tr>";
$i=0;
}
echo '<td><a href="index.php? route=product/manufacturer/info&
manufacturer_id='. $row["manufacturer_id"] .'">. <img src="image/' .
$row["image"] .'"></a></td>';
$i++;
}
} else {
echo "0 results";
}
$conn->close();
I am trying to loop through my database and check to see if the user already exists in another table. If they do then I want to increment a value, if they don't then I want to add the user.
When I run the code below it happily loops through all the results:
<?php
$servername = "p:10*********";
$username = "*******";
$password = "*******";
$dbname = "******";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT * FROM payroll WHERE user != ' ' ";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
echo $result->num_rows;
while($row = $result->fetch_assoc()) {
$user = $row['user'];
$time = $row['time'];
$id = $row['id'];
echo $id;
echo $user;
}
} else {
echo "0 results";
}
$conn->close();
?>
However when I add in the SQL to check to see if they exist in the other table the loop no longer functions correctly and echos the same user each time.
<?php
$servername = "*******";
$username = "******";
$password = "********";
$dbname = "*****";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT * FROM payroll WHERE user != ' ' ";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
echo $result->num_rows;
while($row = $result->fetch_assoc()) {
$user = $row['user'];
$time = $row['time'];
$id = $row['id'];
echo $id;
echo $user;
// Added existing user check:
$sql = "SELECT * FROM smsreport WHERE user = '$user'";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
echo "found";
} else {
echo "USER NOT FOUND";
}
}
} else {
echo "0 results";
}
$conn->close();
?>
In the open eye:
Rename the inside $result variable. It is over writting the first $result.
It could be the problem. Not tested though.