Question

我正在尝试从PHP进行程序化Google搜索，但无济于事。我得到了我所追求的JSON对象，但它有[totalResults] =＆gt; 0，对于我搜索过的所有内容。我使用的apiKey是服务器密钥，customSearchEngineKey用于我通过说明here创建的自定义搜索引擎。我知道这个有用，因为我已经尝试过其他语言，但是不能使用otherLanguage作为最终的程序（我宁愿不说为什么），但是当我把它带到PHP时，由于某种原因，它失败了。为什么？

这是我的PHP代码，顺便说一下：

<?php
    // turn on the use of session variables, if it has not already been done
    session_start();
    // declare function that creates the URL for the Google search associated with a query
    function makeSearchString($query, $startNumber = 1, $count = 10)
    {
        // declare $apiKey,$customSearchEngineKey
        $apiKey = "AIzaSyCW6jCkVPGWRd2PN1ZHeOKq8haJqkYqEwQ";
        $customSearchEngineKey = "003207291839125798740:4fbhl2kr0mi";
        // set up searchString with $apiKey,$customSearchEngineKey
        $searchString = "https://www.googleapis.com/customsearch/v1?key=" . $apiKey . "&cx=" . $customSearchEngineKey . "&q=";
        // split query into an array with ' ' as the delimiter (regex is "/[ ]+/")
        $theQueries = explode("/[ ]+/", $searchString);
        // for each subquery in theQueries
        foreach ($theQueries as $subquery)
            // append subquery to searchString
            $searchString .= $subquery;
        // specify that the response should be in JSON, in searchString
        $searchString .= "&alt=json";
        // try to turn safeSearch on
        $searchString .= "&safe=high";
        // if startNumber is not 1 
        if ($startNumber != 1)
            // specify startNumber, in searchString
            $searchString .= ("&start=" . $startNumber);
        // if count is not 10 
        if ($count != 10)
            // specify count, in searchString
            $searchString .= ("&num=" . $count);
        // return searchString
        return $searchString;
    }

    /* Don't forget that you need to test this function. Try it with at least 2 searches... */
    // declare function that returns the JSON associated with a Google search tied to query
    function getJSONStringFor($query)
    {   
        // if $query is a URL that contains the first few characters in the defaultSearchString
        if (strpos($query, "https://www.googleapis.com/customsearch/v1?key=") !== false)
        {
            // return the file contents for that $query
            return file_get_contents($query);
        }       
        // makeSearchString for $query and get the JSONString from that searchString
        return file_get_contents(makeSearchString($query));
    }

    // if the searchData in $_SESSION is not already initialized
    if (!isset($_SESSION['searchData']))
        // initialize it
        $_SESSION['searchData'] = array();
    // get the terms to search for /* where from, I don't know!! They should at least be an array of Strings */
    /* For now, we can simply create an array of three terms to search for */
    $terms = array("alopecia", "Superman", "Spiderman");
    // for each term
    foreach ($terms as $term)
    {
        // get the JSONString
        // parse the JSONString into an associative array
        $array = json_decode(getJSONStringFor($term), true);
        print_r($array);    /* returns no results for some reason */
        // for each searchResult in that array 

            // if there is pagemap and it has a cse_image 
                // add it as a searchResult for that queryResult
        // append queryResult to searchData
    }
?>

Answer 1

我明白了。问题出在我的makeSearchString（）上，即这一行：

$theQueries = explode("/[ ]+/", $searchString);

当我试图将$searchString拆分成数组时，我不小心告诉PHP编译器将$query拆分成数组。所以，我把它改为$theQueries = explode("/[ ]+/", $query);，一切顺利！

PHP Google搜索没有返回任何内容

1 个答案: