Javascript 如何使用php在链接中使用if-else类

如何使用PHP在链接中使用if-else类

我想在课堂上添加条件我如何才能做到这一点请帮助我解决这个问题

谢谢

if ($user == 'brother' ) {

echo "<img src=\"images/bandmember.jpg\" class=\"right\" >";
} else {
    echo "<img src=\"images/gismember.jpg\" class=\"left\" >";
}

它将显示这样的结果

<img src="images/bandmember.jpg" class="right">
<img src="images/gismember.jpg" class="right">

<img src="images/bandmember.jpg" class="right">
<img src="images/gismember.jpg" class="left">

我想要这样的结果

<img src="images/bandmember.jpg" class="right">
<img src="images/gismember.jpg" class="right">

<img src="images/bandmember.jpg" class="right">
<img src="images/gismember.jpg" class="left">

这是完整的代码

<?php

include 'db.php';


   $sql = 'SELECT * FROM users';

   if($result = mysqli_query($con, $sql)) {
      if(mysqli_num_rows($result) > 0) {


         while($row = mysqli_fetch_array($result)){
            echo "<div class=\"container\">";


              if ($row['username'] == 'brother' ) {

    echo "<img src=\"images/" . $row['photo'] . "\" class=\"right\" >";
    } else {
        echo "<img src=\"images/" . $row['photo'] . "\" class=\"left\" >";
    }

        echo "<span class=\"time-right\">" . $row['username'] . "</span>";

            echo "</div>";
         }
         mysqli_free_result($result);
      } else {
         echo "No records matching your query were found.";
      }
   } else {
      echo "ERROR: Could not able to execute $sql. " . mysqli_error($con);
   }
   mysqli_close($con);
?>

---

您可以使用变量来存储类值，并在第一次迭代后对其进行更改

$class = 'right' ;
while($row = mysqli_fetch_array($result)) {
    echo "<div class=\"container\">";
    if ($row['username'] == 'brother' ) {
        echo "<img src=\"images/" . $row['photo'] . "\" class=\"$class\" >";
    } else {
        echo "<img src=\"images/" . $row['photo'] . "\" class=\"$class\" >";
    }
    $class = 'left'; // change from 'right' to 'left'

或者，您可以使用布尔值并在每次迭代时将其反转。然后使用三元条件创建left或right类：

$right = true ;
while($row = mysqli_fetch_array($result)) {
    echo "<div class=\"container\">";
    if ($row['username'] == 'brother' ) {
        echo "<img src=\"images/" . $row['photo'] . "\" class=\"".($right?'right':'left')."\" >";
    } else {
        echo "<img src=\"images/" . $row['photo'] . "\" class=\"".($right?'right':'left')."\" >";
    }
    $right = !$right; // invert each time

---

试试这个，确保它对你有帮助。 多谢各位

<?php include 'db.php'; $sql = 'SELECT * FROM users'; if($result = mysqli_query($con, $sql)) {
  if(mysqli_num_rows($result) > 0) {


     while($row = mysqli_fetch_array($result)){
        echo "<div class='container'>";


          if ($row['username'] == 'brother' ) {

echo "<img src='images/'" . $row['photo'] . "class='right'>";
} else {
    echo "<img src='images/'" . $row['photo'] . "class='left' >";
}

    echo "<span class='time-right'>" . $row['username'] . "</span>";

        echo "</div>";
     }
     mysqli_free_result($result);
  } else {
     echo "No records matching your query were found.";
  } } else {
  echo "ERROR: Could not able to execute $sql. " . mysqli_error($con);  }  mysqli_close($con); ?>

---

请添加生成实际结果的代码。您的代码一次只打印一个图像的html。添加原始代码并输出。现在你的输出与你的代码不匹配用右替换左，否？我更新完整的代码你想从右到左交替，依此类推？还是先右后左？这是你要找的东西吗？