我如何使用给定的php代码在html页面中显示所有json数据

Question

当我在搜索字段中输入电影名称时，我想显示所有是json文件的电影，但是使用此代码，我只能获得其中一部电影，请您帮助我。

<?php

            if (isset($_POST['submit-search'])) {

                $txtresult = $_POST['search'];

                function    getImdbRecord($title, $ApiKey)
                    {
                        $path = "http://www.omdbapi.com/?s=$title&apikey=$ApiKey";
                        $json = file_get_contents($path);
                        return json_decode($json, TRUE);

                    }
                $data = getImdbRecord($txtresult, "f3d054e8");  


                 echo "<div class = 'info-box'><img src =".$data['Poster']."</img><h3> Name :".$data['Title']."</h3><h3> Year : ".$data['Year']."</h3><h3> Duration : ".$data['Runtime'],"</h3></div>";



    }

        ?>

Answer 1

您需要使用foreach循环来获取所有搜索结果。

喜欢。

$data = getImdbRecord($txtresult, "f3d054e8");  
foreach($data['Search'] as $value){
    echo "<div class = 'info-box'><img src =".$value['Poster']."</img><h3> Name :".$value['Title']."</h3><h3> Year : ".$value['Year']."</h3><h3> Duration : ".$value['Runtime'],"</h3></div>";
}

完整的代码将。

 <?php
 //add function out side the if condition
function getImdbRecord($title, $ApiKey){
                $path = "http://www.omdbapi.com/?s=$title&apikey=$ApiKey";
                $json = file_get_contents($path);
                return json_decode($json, TRUE);

}


if (isset($_POST['submit-search'])) {
   $txtresult = $_POST['search'];
   $data = getImdbRecord($txtresult, "f3d054e8");  
   //use loop to get all the seacrh result.
    foreach($data['Search'] as $value){
        echo "<div class = 'info-box'><img src =".$value['Poster']."</img><h3> Name :".$value['Title']."</h3><h3> Year : ".$value['Year']."</h3><h3> Duration : ".$value['Runtime'],"</h3></div>";
    }
}
?>