使用file_get_contents（）处理上传文件

Question

我有一个处理上传的CSV到临时目录的PHP脚本，然后我有5行代码将CSV转换为JSON。

我的PHP脚本：

if (isset($_FILES['csvList']['type'])) { 

    $validExtensions = array("csv"); 
    $temporary = explode(".", $_FILES["csvList"]["name"]); 
    $file_extension = end($temporary); 

    if (in_array($file_extension, $validExtensions)) { 

    if ($_FILES["csvList"]["error"] > 0) { 

     echo "Return Code: " . $_FILES["csvList"]["error"] . "<br/><br/>"; 
    } else { 

     if (file_exists("/var/www/tmp/" . $_FILES["csvList"]["name"])) { 

     echo $_FILES["csvList"]["name"] . " <span id='invalid'><b>already exists.</b></span> "; 

     } else { 

     $sourcePath = $_FILES['csvList']['tmp_name']; // Storing source path of the file in a variable 
     $targetPath = "/var/www/tmp/".$_FILES['csvList']['name']; // Target path where file is to be stored 

     move_uploaded_file($sourcePath,$targetPath) ; // Moving Uploaded file 

     $csvFile = $sourcePath; 
     $csv = file_get_contents($csvFile); 
     $csvArray = array_map("str_getcsv", explode("\n", $csvFile)); 
     $csvToJson = json_encode($csvArray); 
     print_r($csvToJson); 
     } 
    } 
    } 
} 
$sourcePath = $_FILES['csvList']['tmp_name'];  // Storing source path of the file in a variable 
$targetPath = "/var/www/tmp/".$_FILES['csvList']['name']; // Target path where file is to be stored 
move_uploaded_file($sourcePath,$targetPath) ; // Moving Uploaded file 

的问题是在这条线：print_r($csvToJson);。 这是输出：

[["\/tmp\/phpYeuuBB"]] 

这是我的临时文件的文件路径，我究竟做错了什么？

这里是我的CSV是什么样子 -

CSV Demo

更新：我的JSON是不正确格式化“和\旁边名

{"data":["[[\"Debra Brown\"],[\"Jacqueline Garza\"],[\"Kenneth Foster\"],[\"Antonio Howell\"],[\"Fred Rogers\"],[\"Robert Stanley\"],[\"Jesse Price\"],[\"Henry Bishop\"],[\"Marilyn Phillips\"],[\"Charles White\"],[\"Dennis Lawrence\"],[\"Nicholas Thompson\"],[\"Chris Graham\"],[\"Louis Dean\"],[\"Katherine Green\"],[\"Janice Peters\"],[\"Bobby Wood\"],[\"Bruce King\"],[\"Diane Mills\"],[\"Jane Fields\"],[\"Amanda Gutierrez\"],[\"Russell Cunningham\"],[\"Judith Matthews\"],[\"Carol Franklin\"],[\"Jose Murray\"],[\"Kathryn Cole\"],[\"Katherine Gardner\"],[\"Lois Woods\"],[\"Andrew Bryant\"],[\"Victor Wright\"],[\"Adam Russell\"],[\"Tina Gilbert\"],[\"Shawn Boyd\"],[\"Wanda Porter\"],[\"Rose Morris\"],[\"John Mccoy\"],[\"Frances Gibson\"],[\"Willie Lopez\"],[\"Chris Reyes\"],[\"Craig Vasquez\"],[\"Diane Simmons\"],[\"Mary Little\"],[\"Patricia Fowler\"],[\"Jane Perkins\"],[\"Juan Brooks\"],[\"Bruce Howard\"],[\"Tammy Richardson\"],[\"Jane Gomez\"],[\"Tammy Matthews\"],[\"Matthew Fox\"],[null]]"]} 

应该如何 -

{"data":[["Debra Brown"]]} 

当我打印$csv

Answer 1

看起来你正在输入路径到您的CSV到str_getcsv

$csvArray = array_map("str_getcsv", explode("\n", $csvFile)); 

，而不是实际的CSV内容

$csvArray = array_map("str_getcsv", explode("\n", $csv)); 

Answer 2

您在文件分配给变量$csv但你这样做是：

$csvArray = array_map("str_getcsv", explode("\n", $csvFile)); 

变量$csvFile仍然包含的文件路径您你之前设置为$sourcePath的文件。从file_get_contents()返回的实际文件字符串是$csv。如果将其更改为如下所示：

$csvArray = array_map("str_getcsv", explode("\n", $csv)); 

您可能会发现这解决了您的问题。

这里有the docs为file_get_contents()