Question

我尝试制作一个来自并编写了一个从谷歌获取图片的脚本，但它没有显示任何图像，也没有给出任何错误我应该怎么做。这是我的代码。

<html>
<head>
<title>Images</title>
</head>
<body>
<form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="Post">
<table border="1">
<tr>
<td>Text:</td>
<td><input type="text" name="image" id="ImageType" value=""></td>
<td><input type="Submit" value="Click to Submit"></td>
</tr>
</table>
</form>
<br>

</body>
</html> 

<?php
include "simple_html_dom.php";
if(isset($_POST['submit'])) 
{ 
$search_query = $_POST['image'];


 // $search_query = "love";
 $search_query = urlencode( $search_query );
 $html = file_get_html( "https://www.google.com/search?q=$search_query& tbm=isch" );
 // $image_container = $html->find('div#rcnt', 0);
  $images = $html->find('img');
  $image_count = 10; //Enter the amount of images to be shown
  $i = 0;
  foreach($images as $image){
    if($i == $image_count) break;
    $i++;
    // DO with the image whatever you want here (the image element is '$image'):
    echo $image;
}
}
?>

Answer 1

$html = file_get_html( "https://www.google.com/search?q=$search_query& tbm=isch" );

应该是

$html = file_get_html( "https://www.google.com/search?q=$search_query&tbm=isch" );

需要删除search_query&和tbm之间的空格。

修改

<?php if (isset($_POST['submit'])) { include "simple_html_dom.php"; $search_query = $_POST['image']; $html = file_get_html("https://www.google.com/search?q=$search_query&tbm=isch"); $images = $html->find('img'); $image_count = 10; //Enter the amount of images to be shown $i = 0; foreach ($images as $image) { if ($i == $image_count) break; $i++; // DO with the image whatever you want here (the image element is '$image'): echo $image; } } ?>

编辑 - 2 你应该给name属性提交按钮，如：

<input type="Submit" name="submit" value="Click to Submit">

或使用： if (isset($_POST['image'])) {代替if (isset($_POST['submit'])) {

图像从谷歌提取形式不工作？

1 个答案: