this code works:
$i=1;
require('db.php');
$query = "SELECT own".$i." FROM users WHERE username='".$_SESSION['username']."'";
$result = mysqli_query($con, $query);
$row = mysqli_fetch_assoc($result);
So i try to put it into loop but it is not working. I get only the first result.
Is it even possible to do it that way?
while (isset($row['own'.$i])) {
require('db.php');
$query = "SELECT own".$i." FROM users WHERE username='".$_SESSION['username']."'";
$result = mysqli_query($con, $query);
$row = mysqli_fetch_assoc($result);
echo $row['own'.$i];
$i++;
}
Also I use the above code to make first check before the loop
You are always selecting only SELECT own1.
You need foreach instead of while
require('db.php');
$query = "SELECT * FROM users WHERE username='".$_SESSION['username']."'"; // SQL Injection HERE!
$result = mysqli_query($con, $query);
$rows = mysqli_fetch_assoc($result);
foreach ($rows as $row)) {
var_dump($row);
}
Related
So I need to run my sql query in a foreach loop, but, say there are two variables in the foreach loop, the query only executes the iteration with the first variable twice, instead of executing both the first and the second variable.
My code
$sql = "SELECT * FROM users WHERE idUsers = '$selected';";
$result = mysqli_query($conn, $sql);
if($row = mysqli_fetch_assoc($result))
{
foreach($_POST['order-check'] as $check)
{
$sql2 = "UPDATE order_table SET order_status = 'Processing', assigned_vendor = '$selectedvendor' WHERE order_id = '$check';";
$result2 = mysqli_query($conn, $sql2);
exit();
}
}
else{
echo "failed";
exit();
}
Here, $selected is a POST variable from another page
As Qirel mentioned, remove the "exit()" from your foreach statement.
Also, please ensure you sanitize any POST or GET variables before inserting into the database :)
Your statement should look like this if you want to loop through all $_POST variables
foreach($_POST['order-check'] as $check)
{
$sql2 = "UPDATE order_table SET order_status = 'Processing', assigned_vendor = '$selectedvendor' WHERE order_id = '$check';";
$result2 = mysqli_query($conn, $sql2);
//exit();
}
I need to append result of mysql query with existing result of another query, and both queries performed on different table, actually what I really want to do is take value from main table and get some data from another table using first value, and combine all together using json encode.
$query = "select ID,TIMESTAMP,UUID from TABLE1 where USERNAME='$user'";
$result = mysqli_query($conn, $query);
$numrows = mysqli_num_rows($result);
if($numrows>0)
{
$res=$a;
$myArray = array();
while($row = mysqli_fetch_array($result, MYSQLI_ASSOC)) {
$eachuuid = $row[UUID];
$query1 = "select ID,IMAGEURL from TABLE_IMAGES where IMG_UUID='$eachuuid'";
$result1 = mysqli_query($conn, $query1);
$numrows1 = mysqli_num_rows($result);
if($numrows1>0){
$res1;
$myArray1 = array();
while($row1 = mysqli_fetch_array($result1, MYSQLI_ASSOC)) {
$myArray1[] = $row1;
}
$row['IMG_URL']=$myArray1;
}
$myArray[] = $row;
}
$res['Data']=$myArray;
echo json_encode($res);
Result:
{"Response":"OK","Data":[{"ID":"62",
"TIMESTAMP":"26 January",
"UUID":"12345",
"IMG_URL":[{"ID":"5","IMAGEURL":"26_January_2016_22_39_28_crop.jpg"}]}]}
And I need to get the output like,
{"Response":"OK","Data":[{"ID":"62",
"TIMESTAMP":"26 January",
"UUID":"12345",
"IMG_URL":{"ID":"5","IMAGEURL":"26_January_2016_22_39_28_crop.jpg"}}]}
Means need to remove [] from IMG_URL section so that I can parse the data easily.
You can do this
$row['IMG_URL']=$myArray1[0];
Or
$myArray1 = null;
while($row1 = mysqli_fetch_array($result1, MYSQLI_ASSOC)) {
$myArray1 = $row1;
}
$row['IMG_URL']=$myArray1;
I hope this helps.
How to get total number of row by using this function php mysql ?
i use this code for display data from mysql. It's work good,
Buy i want to know can i get total number of row by using this code ?
<?PHP
include("connect.php");
$query = "SELECT * FROM table";
$result = mysql_query($query) or die(mysql_error());
while($row = mysql_fetch_array($result))
{
$id = $row['id'];
}
?>
Try this mysql_num_rows ($result)
Use this line of code
<?PHP
include("connect.php");
$query = "SELECT * FROM table";
$result = mysql_query($query) or die(mysql_error());
$number_of_rows = mysql_num_rows($result); // this will return the number of rows found by the $result query.
echo $number_of_rows;
while($row = mysql_fetch_array($result))
{
$id = $row['id'];
}
?>
<?php
function comment($postid,$db_con)
{
$commentdiv='';
$sql="SELECT userid,time,comment FROM comments WHERE postid='$postid' LIMIT 3";
$query = mysqli_query($db_con, $sql);
while($row = mysqli_fetch_array($query,MYSQLI_ASSOC)){
$uid=$row["userid"];
$timecomment=$row["time"];
$comment=$row["comment"];
$sql="SELECT username,photo FROM users WHERE id='$uid'";
$query = mysqli_query($db_conx, $sql);
$row = mysqli_fetch_array($query,MYSQLI_ASSOC);
$username=$row["username"];
$photo=$row["photo"];
$userphoto='<img src="xxx/'.$username.'/'.$photo.'">';
if($photo== NULL){
$userphoto = '<img src="xxx/default.png">';
}
$commentdiv.='<div class="xxxxxxx"><div class="yyyyyy">'.$userphoto.'</div><div class="zzzzz">'.$username.'</div><div class="vvvvv">'.$comment.'</div></div>';
}
return $commentdiv;
}
?>
I am new to PHP, I am trying to return 3 comments from above PHP code, but above code returns only 1 row from database, why does fetch array return only 1 row when there is more then 1 row?
Try,I use $query2 = mysqli_query($db_conx, $sql); for your second query,Because it will reset first loop
<?php
function comment($postid,$db_con)
{
$commentdiv='';
$sql="SELECT userid,time,comment FROM comments WHERE postid='$postid' LIMIT 3";
$query = mysqli_query($db_con, $sql);
while($row = mysqli_fetch_array($query,MYSQLI_ASSOC)){
$uid=$row["userid"];
$timecomment=$row["time"];
$comment=$row["comment"];
$sql="SELECT username,photo FROM users WHERE id='$uid'";
$query2 = mysqli_query($db_conx, $sql); // added new variable
$row = mysqli_fetch_array($query2,MYSQLI_ASSOC);
$username=$row["username"];
$photo=$row["photo"];
$userphoto='<img src="xxx/'.$username.'/'.$photo.'">';
if($photo== NULL){
$userphoto = '<img src="xxx/default.png">';
}
$commentdiv.='<div class="xxxxxxx"><div class="yyyyyy">'.$userphoto.'</div><div class="zzzzz">'.$username.'</div><div class="vvvvv">'.$comment.'</div></div>';
}
return $commentdiv;
}
?>
You are overwriting the mysql resource variable that gives the result inside the while loop.
while($row = mysqli_fetch_array($query,MYSQLI_ASSOC)){
^^^^^^ Original resource variable
and inside the while loop again you are using the query variable
$query = mysqli_query($db_conx, $sql);
^^^^^^ Overwriting $query inside the while loop
One suggestion would be to rename the variable inside the loop to something else.
I am creating a function that will retrieve an id number from a table with multiple columns, put it in a variable which i want to use to send as a "claims number" in an email to a claimant and also echo on a confirmation screen. Here's what I have so far.
$sql = "SELECT id FROM warranty_claim where lname=$lname";
$result = mysql_query($sql);
$claim_id = $result;
$result = mysql_fetch_array(mysql_query($sql));
$claim_id = $result['id'];
Supposing you are sure that you will receive 1 row.
$sql = "SELECT id FROM warranty_claim where lname=$lname";
$result = mysql_query($sql);
$c = mysql_fetch_assoc($result);
$claim_id = $c['lname'];
NB*: And get ready for a lecture. Someone is going to tell you that mysql is depracated.
Start using MySQLi or PDO (recommended).
Try :
$sql = "SELECT id FROM warranty_claim where lname=$lname";
$result = mysql_query($sql);
$row = mysql_fetch_row($result);
$claim_id = $row[0];
or
$sql = "SELECT id FROM warranty_claim where lname=$lname";
$result = mysql_query($sql);
while ($row = mysql_fetch_array($result, MYSQL_ASSOC)) {
$claim_id = $row['id'];
}
$row = mysql_fetch_array($result);
$claim_id = $row['id'];
please note that you should also check for errors (e.g. if($result === FALSE)). Also, there are other ways to fetch data - I recommend looking at mysql_fetch_array, mysql_fetch_row, mysql_fetch_assoc and mysql_fetch_object
Also, don't rely on the statement always returning exactly one row - make sure that the result is as expected: mysql_num_rows
Give this a try:
$sql = "SELECT id FROM warranty_claim where lname='".$lname."'";
$result = mysql_query($sql);
$objResult = mysql_fetch_array($result);
$claim_id = $objResult['id'];