小编Ray*_*aga的帖子

Guzzle Error count(): 参数必须是一个数组或一个在

我正在关注下面的文章, https://developers.google.com/analytics/devguides/reporting/core/v3/quickstart/web-php

但是最后,当我尝试使用示例代码(HelloAnalytics.php)时,它会在命令行中显示如下错误并且无法获取数据;

PHP 警告:count(): Parameter must be an array or an object that implementation Countable in C:\xampp\htdocs\vendor\guzzlehttp\guzzle\src\Handler\CurlFactory.php on line 67

警告:count(): Parameter must be an array or an object that implementation Countable in C:\xampp\htdocs\vendor\guzzlehttp\guzzle\src\Handler\CurlFactory.php on line 67 PHP Warning: count(): Parameter must是第 67 行 C:\xampp\htdocs\vendor\guzzlehttp\guzzle\src\Handler\CurlFactory.php 中实现 Countable 的数组或对象

警告:count():参数必须是在 C:\xampp\htdocs\vendor\guzzlehttp\guzzle\src\Handler\CurlFactory.php 中实现 Countable 的数组或对象,第 67 行 PHP 致命错误:未捕获的 Google_Service_Exception:{" error":{"errors":[{"domain":"usageLimits","re​​ason":"accessNotConfigured","message":"Project 687417168367 未找到且不能用于 API 调用。如果它是最近创建的,通过访问https://console.developers.google.com/apis/api/analytics.googleapis.com/overview?project=687417168367启用 Google Analytics API,然后重试。如果您最近启用了此 API,请等待几分钟以进行操作传播到我们的系统并重试。","extendedHelp":" https://console.developers.google。com/apis/api/analytics.googleapis.com/overview?project=687417168367"}],"code":403,"message":"项目 687417168367 未找到,无法用于 API 调用。如果它是最近创建的,请访问https://console.developers.google.com/apis/api/analytics.googleapis.com/overview?project=687417168367启用 Google Analytics …

php google-api google-analytics-api google-api-php-client

3
推荐指数
2
解决办法
6205
查看次数

从foreach循环获取每个数组项作为表数据

我在20个名字的数组上尝试foreach循环.我应该派生一个包含4列和5行的表,每个单元格(表数据)都有一个唯一的名称.我的代码在下面,带有输出表的快照.它还没有奏效.我怎样才能解决这个问题?

<?php

$names = array("Patrick","Raymond","George","Hosea","Samuel","Alan","June","Edwin","Yvonne","John","Paul","Ruto","Uhuru","Raila","Kalonzo","Sonko","Joho","Wetangula","Mudavadi","Matiang'i");


echo "<table width='200' border='1' >";

foreach($names as $name){

echo "<tr>";

    for($cols=1;$cols<5;$cols++){

    echo "<td>".$name."</td>";

    }

echo "<tr>";

}

echo "<table>";

?>
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在此输入图像描述

php loops

-3
推荐指数
1
解决办法
345
查看次数