PHP mysqli_fetch_array 如何转到下一行？

Question

我有一个 php 和 mysql 脚本，我想在其中根据 mysql 表中找到的值创建不同的结果。

我将 horas 数组中的 te 值与该行的 hora 变量的值进行比较。如果我做其他事情，它也会做其他事情。然而我的疑问是，在 if 中，当条件为真时，我如何更改到下一行？

$con = db_connect();


$result = mysqli_query($con,"SELECT * FROM projetor WHERE data = '$mydate' AND hora = '$hora'");

 $horas=array("a","b","c","d","e","f","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v","w","x");

$arrlength=count($horas);

while($row = mysqli_fetch_array($result)){

$cod = $row['cod'];
    $data = $row['data'];
    $hora = $row['hora'];

    $professor = $row['professor'];
    $sala = $row['sala'];


    for($x=0;$x<$arrlength;$x++)
      {

          if ($horas[$x] == $hora){

                  echo "<tr>";

                  echo "<td>" .$professor . "</td>";

                  echo "</tr>";

              next($row);//how to go to next row??
          }
          else{

              $h=$horas[$x];
              $professor ="<a href='add_reserva.php?hora=$h&mydate=$mydate'>Requisitar</a>";
              echo "<tr>";
              echo "</tr>";


              }
      }

  }

Answer 1

您不需要调用任何 next 方法，此代码会为您完成此操作:

while($row = mysqli_fetch_array($result)){

(已编辑)由于您的 while 循环中有一个 for 循环迭代，因此您可以放置​​一个 break; 来结束您的内部 for 循环，这将开始 while 循环的下一次迭代(这将转到 SQL 结果的下一行)