Question

我试图从mysql中的数据库中获取图像，然后在3行等中水平显示它们，我还希望将相关图像显示在图像的顶部。我不知道该怎么做。在我的主要部分中看到我的代码。

beforeEach(() => {
    addProviders([
        {provide: Router, useClass: MockRouter}
    ])
});

这是我的 php查询，我觉得这很好。

<div class="container" id="content">
    <!-- Example row of columns -->
    <div class="row">
        <div class="col-md-4">
               <?php do { ?>              

            <a href="missions.php?missions_id=<?php echo $rsMissions['missions_id']; ?>">
            <img src="<?php echo $rsMissions['missions_image']; ?>" width="500px" height="350px" >
            <p><?php echo $rsMissions['missions_name']; ?></p>
            </a>

<?php } while ($rsMissions = mysqli_fetch_assoc($missions_query)) ?>
</div>
</div>

以下是相关div的 css

<?php
  require_once('includes/dbconn.php');
  $missions_sql = "SELECT missions_id, missions_name, missions_image FROM missions";
$missions_query = mysqli_query($dbconn, $missions_sql) or die(mysqli_error());
$rsMissions = mysqli_fetch_assoc($missions_query);

?>

Answer 1

你必须把循环放在col-md-4类之前。否则div不会重复。你不能水平地获取你的图像。尝试下面的代码，它会很适合你。

<div class="container" id="content">
<!-- Example row of columns -->
<div class="row">
  <?php do { ?>
    <div class="col-md-4">
        <a href="missions.php?missions_id=<?php echo $rsMissions['missions_id']; ?>">
        <img src="<?php echo $rsMissions['missions_image']; ?>" width="500px" height="350px" >
        <p><?php echo $rsMissions['missions_name']; ?></p>
        </a>
    </div>
 <?php } while ($rsMissions = mysqli_fetch_assoc($missions_query)) ?>

Answer 2

我认为你把代码放在了错误的地方。如果有效，请尝试更改下面的代码。

function handleKeydown(e) {
        // Allow: backspace, delete, tab, escape, enter and .
        if ($.inArray(e.keyCode, [46, 8, 9, 27, 13, 110]) !== -1 ||
            // Allow: Ctrl+A, Command+A
            (e.keyCode == 65 && ( e.ctrlKey === true || e.metaKey === true ) ) || 
            // Allow: home, end, left, right, down, up
            (e.keyCode >= 35 && e.keyCode <= 40)) {
            // let it happen, don't do anything
            return;
        }
        // Ensure that it is a number and stop the keypress
        if ((e.shiftKey || (e.keyCode < 48 || e.keyCode > 57)) && (e.keyCode < 96 || e.keyCode > 105)) {
            e.preventDefault();
        }
    }

$('#content')
.html('Label 1: <input name="quantity[]" class="quantity"> Label 2: <input name="quantity[]" class="quantity">')
.find(".quantity")
.keydown(handleKeydown);

Answer 3

<?php
$servername = 'localhost';
$username = 'root';
$password = '';
$dbname = 'ledp';

//create connection
$conn = new mysqli($servername,$username,$password,$dbname);

//check connection
if($conn->connect_error) {
    die ('Error: Failed to connect database'.$conn->connect_error);
}

if(isset($_POST['submit5'])) {
    $pro_image = $_FILES['pro_image'];
    
    $tmp = $_FILES['pro_image']['tmp_name'];
    $name = $_FILES['pro_image']['name'];

    move_uploaded_file($tmp,'upload/'.time().$name);

    if($name = '') {
        echo 'Image upload failed';
    } else {
        echo 'Image uploaded';
    }
}


$files = glob('upload/*.*');
//loop
$i;
for($i=0;$i<count($files);$i++){
    $image = $files[$i];
    echo basename($image);
    echo '<div class="container-fluid">';
        echo '<div class="row">';
            echo '<div class="col-3 d-flex">';
                echo '<img class="img-fluid" src="'.$image.'" alt="" width = 300px>'.'<br><br>';
            echo '</div>';
        echo '</div>';
    echo '</div>';
};
 
echo "<h2>Total images: $i </h2>";

$conn->close();

?>

如何使用PHP和mysql水平显示图像？

3 个答案: