I am trying to echo the data once in one field. However it is looping and i'm not sure what to change to make it echo once. Does anyone know how? Here is the code:
<?php
$email = $_POST['email'];
$servername = "localhost";
$username = "root";
$password = "";
$dbName = "users";
//Make Connection
$conn = new mysqli($servername, $username, $password, $dbName);
//Check Connection
if(!$conn){
die("Connection Failed. ". mysqli_connect_error());
}
$sql = "SELECT `profilepicurl` FROM users WHERE `email`='".$email."'";
$result = mysqli_query($conn ,$sql);
if(mysqli_num_rows($result) > 0){
//show data for each row
while($row = mysqli_fetch_assoc($result)){
echo $row['profilepicurl'];
}
}
?>
Thanks!
This is quite simple!
Change while($row = mysqli_fetch_assoc($result)){ to $row = mysqli_fetch_assoc($result){
Remove the while() loop and the braces:
if ( mysqli_num_rows($result) > 0 ) {
//show data for each row
$row = mysqli_fetch_assoc($result);
echo $row['profilepicurl'];
}
The correct way to stop a loop is using break:
if(mysqli_num_rows($result) > 0){
//show data for each row
while($row = mysqli_fetch_assoc($result)){
echo $row['profilepicurl'];
break;
}
}
Although, if you just need 1 value, you can remove the while loop and use only:
$row = mysqli_fetch_assoc($result);
echo $row['profilepicurl'];
Notes:
Your script is unsafe due to the possibility of sql injection, make sure you read how can i prevent sql injection in php.
Related
I'm trying to convert the result that i'm getting from mysql to a php array
can anyone helps me
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "women";
$conn = new mysqli($servername, $username, $password, $dbname);
$id=$_GET['id'];
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT DAY(ADDDATE(`dateDebutC`, `dureeC`)) AS MONTHS,
DAY(ADDDATE(ADDDATE(`dateDebutC`, `dureeC`),`dureeR`))AS DAYS
FROM normalW
where id = '$id'";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
foreach($new_array as $array){
echo $row['DAYS'].'<br />';
echo $row['MONTHS'].'<br />';
}
} else {
echo "0 results";
}
$conn->close();
?>
Problem solved Thank you guys
To answer your question you must first declare the new array
$new_array = array();
Then loop through your query results to populated the array
while ($row = $result->fetch()) {
$new_array[] = $row;
}
But as one of the comments mentioned you really should be using prepared statements to protect yourself from sql injection.
$stmt = $mysqli->prepare("SELECT DAY(ADDDATE(`dateDebutC`, `dureeC`)) AS MONTHS, DAY(ADDDATE(ADDDATE(`dateDebutC`, `dureeC`),`dureeR`)) AS DAYS FROM normalW where id = ?");
/* bind parameters i means integer type */
$stmt->bind_param("i", $id);
$stmt->execute();
$new_array = array();
while($row = $stmt->fetch()) {
$new_array[] = $row;
}
I'm using this code here
<?php
error_reporting(1);
$servername = '127.0.0.1';
$username = '';
$password = '';
$dbname = 'splafpoo_users';
$conn = new mysqli($servername, $username, $password, $dbname);
if (mysqli_connect_errno()){
printf("<b>Connection failed:</b> %s\n", mysqli_connect_error());
exit;
}
$key = '';
if(isset($_POST['key'])){
$key = $_POST['key'];
}
$query = "SELECT * FROM users WHERE serial='$key'";
echo $query;
$result = $mysqli->query($query);
$row = $result->fetch_assoc();
echo $row;
?>
Running the query SELECT * FROM users WHERE serial='test' in phpMyAdmin returns the desired result however when trying to display the result using the code above nothing is displayed and I cannot figure out how. How do I display the result?
You're gonna need a good old fashion while loop
while($row = $result->fetch_assoc()) {
echo $row['WHATEVERCOLUMNITISYOUWANT'];
}
also this is most definitely a duplicate.
Use var_dump($row) instead of echo $row or you use echo with a key:e.g. echo $row["user"]
I have a database which has a table called 'propImages' and there are two columns.- 'pid' and 'location'.
And i have data in the database where multiple images can contained by single pid.
image contains database data
now i want to retrieve images from database according to given pid. there can be more than one image.
All i know it there should be an iteration to retrieve images.
I want to display images in HTML .
can you please show me the way to do it in php?
Thanks in advance guys
This may help you
<?php
include 'inc/database.php';
$conn = new mysqli($servername, $username, $password, $database);
$propid = $_GET['propid'];
$sql = "SELECT * FROM propImages WHERE propid='" . $propid . "';";
$result = $conn->query($sql);
if($result->num_rows > 0) {
while($row = $result->fetch_assoc()) {
echo "<img src=" . $row['image'] . ">";
}
}
else {
echo "No results";
}
?>
in the inc/database.php :
<?php
$servername = "localhost";
$username = "root";
$password = "";
$database = "database";
?>
To see how it works try visiting : file.php?propid=22
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "databasename";
// Create connection
$con = mysqli_connect($servername, $username, $password, $dbname);
//create sql
$sql = "SELECT * FROM `propImages` where pid='$YOUR_PID'";
$result = mysqli_query($con, $sql);
$row = mysqli_num_rows($result);
//retrive data print here
if($row > 0){
while($col = mysqli_fetch_assoc($result))
{
echo $col['location'];
}
} else {
echo 'no result found.';
}
?>
wish it helps
I have a table. In that table, there is are two fields called to and amount. I need to display the sum of amount of all rows where to value is equal to a particular value. (Say 34). How to achieve this?
Code I have done so far
$result = mysql_query("SELECT * FROM transactions WHERE to = '34'")or die(mysql_error());
while($row = mysql_fetch_array( $result )) { echo $row['amount']; } ?>
The above code gives the individual amount value of each row. But what I want is the sum of these values.
The Database
Use SUM to add all the amounts and echo it
$result = mysql_query("SELECT SUM(amount) as amounts FROM transactions WHERE to = '34'")or die(mysql_error());
while($row = mysql_fetch_array( $result )) { echo $row['amounts']; } ?>
I suggest you to use mysqli_* or PDO because mysql_* is deprecated and not available in PHP 7.
Here is the complete example of your code by using MYSQLi Object Oriented:
$servername = "localhost";
$username = "username";
$password = "password";
$dbname = "myDB";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT SUM(amount) as amounts FROM transactions WHERE to = 34";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
echo $row['amounts']. "<br>";
}
}
else
{
echo "0 results";
}
$conn->close();
?>
I believe my PHP to be functioning perfectly, therefore I think it's a query error. When I proceed, with form details stored in the session... it happily returns my Posted information but doesn't seem to be pulling anything from my database - there is a row in my database containing the email address I am using. Does anybody see anything blatantly wrong with this PHP?
Thanks for your help.
<?php
session_start();
$servername = "localhost";
$username = "privatedbroot";
$password = "not4ulol";
$dbname = "pdb_inventory";
$status = $_GET["action"];
$_SESSION["Cemail"] = $_POST["CEMAIL"];
$_SESSION["Access"] = md5($_POST["ACCESS"]);
$conn = new mysqli($servername, $username, $password, $dbname);
$sql = "SELECT CEMAIL, ACCESS FROM POPU WHERE `CEMAIL`= ".$_SESSION['Cemail'];
echo $sql;
$result = $conn->query($sql);
if ($result->num_rows > 0) {
while($row = $result->fetch_assoc()) {
if ($_SESSION["Access"] == $row["ACCESS"]){
echo "password correct!";
} else {
echo "password wrong!";
}
}
}else{
echo "ur email is wrong m8.";
}
?>
Try this:
$cemail = $_SESSION['Cemail'];
$sql = "SELECT CEMAIL, ACCESS FROM POPU WHERE `CEMAIL`= '$cemail'";