Question

我将一些图像存储在数组中并显示为每次刷新时图像被洗牌。现在我想存储我使用php从屏幕中选择的图像。请帮帮我。这是代码

<?php
$images = array(
'<a href='#'><img src="images/images (4).jpg" alt="" width="234" height="212" /></a>',
'<a href='#'><img src="images/images (6).jpg" alt="" width="234" height="212" /></a>',
'<a href='#'><img src="images/images (5).jpg" alt="" width="234" height="212" /></a>', 
'<a href='#'><img src="images/drt.jpg" alt="" width="234" height="212" /></a>',
'<a href='#'><img src="images/rf.jpg" alt="" width="234" height="212" /></a>', 
'<a href='#'><img src="images/yu.jpg" alt="" width="234" height="212" /></a>',
'<a href='#'><img src="images/ed.jpg" alt="" width="234" height="212" /></a>');
shuffle($images); // Randomize images array;
?>
<?php echo $images[0];?>
<?php echo $images[1];?>
<?php echo $images[2];?>

Answer 1

尝试这种方式：

$images = array("img1.jpg","img2.jpg","img3.jpg","img4.jpg");
shuffle($images); // for randomaize
    //save  $images[0], $images[1],  into  database here

    // and echo here..
echo "<img src=\"images/$images[0]\" alt=\"\" width=\"234\" height=\"212\" />\n";
echo "<img src=\"images/$images[0].jpg\" alt=\"\" width=\"234\" height=\"212\" />\n";

Answer 2

您可以使用复选框将每个img打包在标签中，并像简单的html表单一样发送选择。但在此之前，你必须以某种方式识别图像，即图像文件名。例如：

＆＃13;

.image-select-form input[type=checkbox] {
  position:absolute;
  visibility: hidden;
}



.image-select-form label {
  cursor: pointer;
  position: relatice;
  display: inline-block;
  border: 2px solid rgba(0, 0, 0, 0)
}

.image-select-form label img{
  display: block;
}

.image-select-form input[type=checkbox]:checked + label {
  border-color: red;
}

＆＃13;

<form action="select.php" method="post" class="image-select-form">
    <div>
        <input type="checkbox" name="selected_images" value="rf.jpg" checked id="img1">
      <label for="img1"><img src="images/rf.jpg" alt="" width="234" height="212" /></label>
      <input type="checkbox" name="selected_images" value="drt.jpg" id="img2">
      <label for="img2"><img src="images/drt.jpg" alt="" width="234" height="212" /></label>
      <input type="checkbox" name="selected_images" value="yu.jpg" id="img3">
      <label for="img3"><img src="images/yu.jpg" alt="" width="234" height="212" /></label>
      <input type="checkbox" name="selected_images" value="ed.jpg" id="img4">
      <label for="img4"><img src="images/ed.jpg" alt="" width="234" height="212" /></label>
    </div>
    
    <input type="submit" value="Submit selected images">
</form>

＆＃13;

在处理表单的php中，select.php只访问存储所选图像文件名的$_POST['selected_images']

如何在php中存储来自数据库的点击图像？

2 个答案: