坚持一个简单的代码
Stuck on a simple code
我有一个用于打开和读取 csv 文件的代码，它是 fopen() 方法。
但它显示错误。下面是代码
<?php
$file = fopen('C:\Users\Abby\Desktop\testbad\testbed_information.csv', `r`);
?>
错误是
Warning: fopen(C:\Users\Abby\Desktop\testbad\testbed_information.csv): failed to open stream: No error in C:\wamp64\www\date.php on line 2
请帮帮我...!!?
您对报价有疑问。您在第二个参数中使用反引号。用双引号或单引号更改它
<?php
$file = fopen('C:\Users\Abby\Desktop\testbad\testbed_information.csv', "r");
?>
我有一个用于打开和读取 csv 文件的代码，它是 fopen() 方法。 但它显示错误。下面是代码
<?php
$file = fopen('C:\Users\Abby\Desktop\testbad\testbed_information.csv', `r`);
?>
错误是
Warning: fopen(C:\Users\Abby\Desktop\testbad\testbed_information.csv): failed to open stream: No error in C:\wamp64\www\date.php on line 2
请帮帮我...!!?
您对报价有疑问。您在第二个参数中使用反引号。用双引号或单引号更改它
<?php
$file = fopen('C:\Users\Abby\Desktop\testbad\testbed_information.csv', "r");
?>