如何从数据库中回显具有特定ID的行

Question

So first off my database table is set up like this:

id | userid | amount | date | status

1 | 10 | 25.00 | 2017-09-12 | None

and I want to to echo out all the rows that include userid 10 into a html table. I have tried this:

<?php
$id = $_GET['id'];
$userinfo= $mysqli->query("SELECT * FROM invoices WHERE userid = $id");
$row = mysql_fetch_array($userinfo) or die();

echo '<table>';
while($row = mysql_fetch_array($query)){
  echo '<tr>
  <td><font size="2" face="Lucida Sans Unicode" color=#EBEBEB>' .$row['id'].'</td>
  <td><font size="2" face="Lucida Sans Unicode" color=#EBEBEB>' .$row['amount'].'</td>
  <td><font size="2" face="Lucida Sans Unicode" color=#EBEBEB>' .$row['date'].'</td>
  <td><font size="2" face="Lucida Sans Unicode" color=#EBEBEB>' .$row['status'].'</td>
        </tr>';
}
echo '</table>';
?>

and nothing shows up.

Answer 1

Use below code:

1 . use $userinfo instead of $query while($row = mysql_fetch_array($query))

2 . Change $mysqli->query to mysqli_query .

3 . Use mysqli instead of mysql .

<?php
$id = $_GET['id'];

//mysqli_connect("host","username","password","dbname") 
$conn=mysqli_connect("localhost","root","","dbname");
$sql="SELECT * FROM invoices WHERE userid = '.$id.'";
$result=mysqli_query($conn,$sql);

echo '<table>';
while($row = mysqli_fetch_array($result)){
  echo '<tr>
  <td>' .$row['id'].'</td>
  <td>' .$row['amount'].'</td>
  <td>' .$row['date'].'</td>
  <td>' .$row['status'].'</td>
        </tr>';
}
echo '</table>';
?>

Answer 2

you mixing mysqli with mysql and object oriented style with Procedural

use object oriented as below

<?php
$id = $_GET['id'];
$userinfo= $mysqli->query("SELECT * FROM invoices WHERE userid = $id");  
echo '<table>';

while($row = $userinfo->fetch_assoc()){
  echo '<tr>
  <td><font size="2" face="Lucida Sans Unicode" color=#EBEBEB>' .$row['id'].'</td>
  <td><font size="2" face="Lucida Sans Unicode" color=#EBEBEB>' .$row['amount'].'</td>
  <td><font size="2" face="Lucida Sans Unicode" color=#EBEBEB>' .$row['date'].'</td>
  <td><font size="2" face="Lucida Sans Unicode" color=#EBEBEB>' .$row['status'].'</td>
        </tr>';
}
echo '</table>';
?>

Answer 3

Try this instead:

<?php
   $id = $_GET['id'];
   $mysqli = new \mysqli('localhost', 'username', 'password', 'database');
   $userinfo= $mysqli->query("SELECT id, amount, date, status FROM invoices WHERE userid = $id");

  echo '<table>';
  if ($userinfo->num_rows > 0) {
     while($row = $userinfo->fetch_assoc()) {
        echo '<tr>
           <td><font size="2" face="Lucida Sans Unicode" color=#EBEBEB>' .$row['id'].'</td>
           <td><font size="2" face="Lucida Sans Unicode" color=#EBEBEB>' .$row['amount'].'</td>
           <td><font size="2" face="Lucida Sans Unicode" color=#EBEBEB>' .$row['date'].'</td>
           <td><font size="2" face="Lucida Sans Unicode" color=#EBEBEB>' .$row['status'].'</td>
        </tr>';
     }
  }
  else {
     echo "0 rows returned";
  }
  echo '</table>';
?>

Also just me being annoying but don't use * in your select statement. Always list the fields even if you want every single one. You should also be using prepared statements.