Question

我在html输出中遇到了最奇怪的时间。如果你看看我的问题开始的//开始画廊行，第一个输出工作正常。

这是输出的外观

<div class='row'>
    <div class='col-md-12'>
        <div id='gallery-slider' class='slider responsive'>
            <img class='img-responsive' src='cdn/assets/gallery/1.jpg'>
            <img class='img-responsive' src='cdn/assets/gallery/3.jpg'>
            <img class='img-responsive' src='cdn/assets/gallery/2.jpg'>
            <img class='img-responsive' src='cdn/assets/gallery/4.jpg'>
            <img class='img-responsive' src='cdn/assets/gallery/5.jpg'>
        </div>
    </div>
</div>

在//画廊的开头，当我查看来源时，这是输出

<div class='row'>
    <div class='col-md-12'>
        <div id='gallery-slider' class='slider responsive'>
        <img class='img-responsive' src='cdn/assets/gallery/1.jpg'></div>
    </div>
</div>
<img class='img-responsive' src='cdn/assets/gallery/3.jpg'>
<img class='img-responsive' src='cdn/assets/gallery/2.jpg'>
<img class='img-responsive' src='cdn/assets/gallery/4.jpg'>
<img class='img-responsive' src='cdn/assets/gallery/5.jpg'>

但无论我把最后一个输出DIV放在哪里都会导致问题

<?php 

$stmt = $db->prepare("my query");
$stmt->execute();
$result = $stmt->get_result();

$output = "";
$checker = [];

while ($row = mysqli_fetch_assoc($result)) {
    $ID = $row['ID'];        
    $FullName = $row['FullName'];    
    $Email = $row['Email'];   
    $JobTitle = $row['JobTitle'];
    $Bio = $row['Bio'];
    $Photo = $row['Photo'];  

    $GalleryImage = explode(',', $row['GalleryImage']);

    if (isset($Photo) && ! empty($Photo)) {
        $ProfileImage = "$Photo";
    } else {
        $ProfileImage= "avatar.jpg";
    }       

    if(!in_array($row['ID'], $checker)) {
        $output .= "
            <div class='container yep team-wrap'>
              <div class='row'>        
                <div class='col-md-6'>
                    <img class='img-responsive' src='cdn/assets/artist/$ProfileImage'>
                </div>
            <div class='col-md-6'>
              <strong>$FullName<br>$JobTitle</strong>
                 <br>
                <p>$Bio</p>
               <a href='mailto:$Email' class='btn btn-info'>Contact Me</a>
            </div> 
        </div>";

        //End of info row

        $output .="<br /><br /><br />";

        //Start Gallery Row
        $output .= "
                <div class='row'>
                 <div class='col-md-12'>
                <div id='gallery-slider' class='slider responsive'>
               ";
    }

    foreach ($GalleryImage as $img){
        //Display this row as many times as needed by data in this row.
        $output .= "<img class='img-responsive' src='cdn/assets/gallery/$img'>";                     
    }
    $output .= "
                    </div>
                </div>
            </div>";
    // End gallery row   
    array_push( $checker, $row['ID']); 
}  

$output .= "</div>";

echo $output;


?>

SQL

$stmt = $db->prepare("
SELECT U.ID,
       U.FullName,
       U.Email,
       U.JobTitle,
       U.Bio,
       U.Photo, G.GalleryImage
FROM users U 
LEFT join gallery G 
ON U.ID = G.ID
 ");


$stmt->execute();
$result = $stmt->get_result();

Answer 1

好的，所以我认为最好的做法是循环你的$ result并过滤掉所有重复的值，并将图像分配给一个以[[ID]]行为键的数组然后循环在检查之后他们！

$checker = array();
$profileArray = array();
while ($row = mysqli_fetch_assoc($result))
{
    if($row['GalleryImage'])
    {
        $profileArray[$row['ID']]['GalleryImages'][] = $row['GalleryImage'];
    }

    if(!in_array($row['ID'], $checker))
    {
        while (list ($key, $value) = each($row))
        {
            if($key != 'GalleryImage')
            {
                $profileArray[$row['ID']][$key] = $value;
            }
        }
        $checker[] = $row['ID'];
    }
}

foreach ($profileArray as $row)
{
    $ID = $row['ID'];
    $FullName = $row['FullName'];
    $Email = $row['Email'];
    $JobTitle = $row['JobTitle'];
    $Bio = $row['Bio'];
    $Photo = $row['Photo'];
    $GalleyImages = $row['GalleryImages'];

    if (isset($Photo) && !empty($Photo))
    {
        $ProfileImage = "$Photo";
    }
    else
    {
        $ProfileImage = "avatar.jpg";
    }

    $output .= "
    <div class='container yep team-wrap'>
      <div class='row'>        
        <div class='col-md-6'>
            <img class='img-responsive' src='cdn/assets/artist/$ProfileImage'>
        </div>
    <div class='col-md-6'>
      <strong>$FullName<br>$JobTitle</strong>
         <br>
        <p>$Bio</p>
       <a href='mailto:$Email' class='btn btn-info'>Contact Me</a>
    </div> 
</div>";

    //End of info row
    $output .= "<br /><br /><br />";

    //Start Gallery Row
    $output .= "
        <div class='row'>
         <div class='col-md-12'>
        <div id='gallery-slider' class='slider responsive'>";

    if(!$GalleyImages)
    {
        foreach ($GalleyImages as $img)
        {
            //Display this row as many times as needed by data in this row.
            $output .= "<img class='img-responsive' src='cdn/assets/gallery/$img'>";
        }
    }
    else
    {
        $output .= "HTML THAT YOU WANNA DISPLAY instead of images";
    }

    $output .= "
                </div>
            </div>
        </div>
    </div>";
}
echo $output;

Answer 2

好的，首先，正确格式化代码总是好的，这样你就不会犯这些错误。

您的第一个输出缺少结束div

<div class='container yep team-wrap'>
    <div class='row'>        
        <div class='col-md-6'>
            <img class='img-responsive' src='cdn/assets/artist/$ProfileImage'>
        </div>

        <div class='col-md-6'>
            <strong>$FullName<br>$JobTitle</strong>
            <br>
            <p>$Bio</p>
            <a href='mailto:$Email' class='btn btn-info'>Contact Me</a>
        </div> 
    </div>
</div> <!-- This was missing-->

最后你已经关闭了你的if语句，快速绕过下面的代码：

//Start Gallery Row
    $output .= "
        <div class='row'>
         <div class='col-md-12'>
        <div id='gallery-slider' class='slider responsive'>
       ";
} // CLOSED AT THE WRONG SPOT

尝试以下方法：

<?php
    $stmt = $db->prepare("my query");
    $stmt->execute();
    $result = $stmt->get_result();

    $output = "";
    $checker = [];

    while ($row = mysqli_fetch_assoc($result)) {
        $ID = $row['ID'];        
        $FullName = $row['FullName'];    
        $Email = $row['Email'];   
        $JobTitle = $row['JobTitle'];
        $Bio = $row['Bio'];
        $Photo = $row['Photo'];  


        $GalleryImage = explode(',', $row['GalleryImage']);


        if (isset($Photo) && ! empty($Photo)) {
            $ProfileImage = "$Photo";
        } else {
            $ProfileImage= "avatar.jpg";
        }       


        if(!in_array($row['ID'], $checker)) {

            $output .= "
                <div class='container yep team-wrap'>
                    <div class='row'>        
                        <div class='col-md-6'>
                            <img class='img-responsive' src='cdn/assets/artist/$ProfileImage'>
                        </div>

                        <div class='col-md-6'>
                            <strong>$FullName<br>$JobTitle</strong>
                            <br>
                            <p>$Bio</p>
                            <a href='mailto:$Email' class='btn btn-info'>Contact Me</a>
                        </div> 
                    </div>
                </div> <!-- This was missing-->
            ";

           //End of info row


            $output .="<br /><br /><br />";


           //Start Gallery Row

           $output .= "
                <div class='row'>
                    <div class='col-md-12'>
                        <div id='gallery-slider' class='slider responsive'>
            ";

            foreach ($GalleryImage as $img) {
               //Display this row as many times as needed by data in this row.
               $output .= "<img class='img-responsive' src='cdn/assets/gallery/$img'>";                     
            }

            $output .= "
                        </div>
                    </div>
                </div>
            ";

            // End gallery row   
            array_push( $checker, $row['ID']); 
        }
    }  


    $output .= "</div>";

    echo $output;
?>

Answer 3

大部分时间都会尝试将PHP与HTML分开，以便您轻松查看错误。

$stmt = $db->prepare("query");
$stmt->execute();
$result = $stmt->get_result();

$output = "";
$checker = [];

while ($row = mysqli_fetch_assoc($result)) {
    $ID = $row['ID'];
    $FullName = $row['FullName'];
    $Email = $row['Email'];
    $JobTitle = $row['JobTitle'];
    $Bio = $row['Bio'];
    $Photo = $row['Photo'];

    $GalleryImage = explode(',', $row['GalleryImage']);

    if (isset($Photo) && !empty($Photo)) {
        $ProfileImage = "$Photo";
    } else {
        $ProfileImage = "avatar.jpg";
    }

    if (!in_array($row['ID'], $checker)) : ?>
        <div class='container yep team-wrap'>
            <div class='row'>
                <div class='col-md-6'>
                    <img class='img-responsive' src='cdn/assets/artist/<?php echo $ProfileImage; ?>'>
                </div>
                <div class='col-md-6'>
                    <strong><?php echo $FullName; ?><br><?php echo $JobTitle; ?></strong>
                    <br>
                    <p><?php echo $Bio; ?></p>
                    <a href='mailto:$Email' class='btn btn-info'>Contact Me</a>
                </div>
            </div>

            <!--        End of info row-->

            <br/><br/><br/>

            <!--        Start Gallery Row-->
            <div class='row'>
                <div class='col-md-12'>
                    <div id='gallery-slider' class='slider responsive'>
                        <!-- Display this row as many times as needed by data in this row.-->
                       <?php foreach ($GalleryImage as $img) : ?>
                        <img class='img-responsive' src='cdn/assets/gallery/<?php echo $img; ?>'>
                        <?php endforeach; ?>
                    </div>
                </div>
            </div>
        </div>
            <!--        // End gallery row-->
        <?php array_push($checker, $row['ID']); endif;
}
?

示例输出

输出html div冲突

3 个答案: