在同一mysqli_fetch_assoc接一个地循环时执行两个循环

Question

请帮助我，一个接一个地循环时，我该如何运行两个。我想获得所有数据库行。如果我将手动控制div放在循环时将手动控制div放置在内部，则会单独显示整个div。当将其放在外面时，它无济于事。请重写代码。

我想像this_一样获得价值，

<div class="img-slider-container">      
<div class="img-slide active">
   <img src="img/1.jpg" alt="Image">
   <div class="img-info">
      <h2>Heading 1</h2>
      <p>Image Text 1</p>
    </div>
 </div>

<div class="img-slide">
   <img src="img/2.jpg" alt="Image">
   <div class="img-info">
      <h2>Heading 2</h2>
      <p>Image Text 2</p>
    </div>
 </div> <!--Continue-->

 <div class="manual-control">
    <div class="slide-btn 1"></div>
    <div class="slide-btn 2"></div> <!--Continue-->
 </div>

我的代码示例被bellow_ _，

<div class="img-slider-container">
        <?php
       $sql = "SELECT * FROM $tbImadeSlide"; 
        $stmt = mysqli_stmt_init($conn);
        if (!mysqli_stmt_prepare($stmt, $sql)) {
            header("Location: index.php?error=sqlerror");
            exit();
        }
        else{
            $result = mysqli_query($conn, $sql);
            $resultCheck = mysqli_num_rows($result);

            if ($resultCheck > 0){
                while($row = mysqli_fetch_assoc($result)){
                    echo '

                        <div class="img-slide '.$row["imageStatus"].'">
                            <img src="img/'.$row["imageName"].'" alt="Image">
                            <div class="img-info">
                                <h2>'.$row["imageHeader"].'</h2>
                                <p>'.$row["imageText"].'</p>
                            </div>
                        </div>';
                       
                     
                } 
                    echo '<div class="manual-control">';
                    while($row = mysqli_fetch_assoc($result)){
                        echo '<div class="slide-btn '.$row["imageId"].'"></div>';}
                    echo '</div>';
            } else{
                echo "<div class='no_result'>No Photos To Slide</div>";
            }
        }
    ?>

但是代码只是在循环this_时首先重复

<div class="img-slider-container">      
    <div class="img-slide active">
       <img src="img/1.jpg" alt="Image">
       <div class="img-info">
          <h2>Heading 1</h2>
          <p>Image Text 1</p>
        </div>
     </div>

    <div class="img-slide">
       <img src="img/2.jpg" alt="Image">
       <div class="img-info">
          <h2>Heading 2</h2>
          <p>Image Text 2</p>
        </div>
     </div> <!--Continue-->

     <div class="manual-control">
        <!--Nothing Shows-->
     </div>
 </div>

Answer 1

如果不保留数组中的所有行，就可以通过 foreach 循环来执行此操作。

只需替换

而（$ row = mysqli_fetch_assoc（$ result））{ foreach（$ resud as $ row）{

您的代码可能看起来像：

<div class="img-slider-container">
    <?php
   $sql = "SELECT * FROM $tbImadeSlide"; 
    $stmt = mysqli_stmt_init($conn);
    if (!mysqli_stmt_prepare($stmt, $sql)) {
        header("Location: index.php?error=sqlerror");
        exit();
    }
    else{
        $result = mysqli_query($conn, $sql);
        $resultCheck = mysqli_num_rows($result);

        if ($resultCheck > 0){
            foreach($result as $row){
                echo '

                    <div class="img-slide '.$row["imageStatus"].'">
                        <img src="img/'.$row["imageName"].'" alt="Image">
                        <div class="img-info">
                            <h2>'.$row["imageHeader"].'</h2>
                            <p>'.$row["imageText"].'</p>
                        </div>
                    </div>';


            } 
                echo '<div class="manual-control">';
                foreach($result as $row){
                    echo '<div class="slide-btn '.$row["imageId"].'"></div>';}
                echo '</div>';
        } else{
            echo "<div class='no_result'>No Photos To Slide</div>";
        }
    }
?>