使用 file_get_contents() 处理上传文件

Question

我有一个 PHP 脚本可以将上传的 CSV 处理到一个临时目录中，然后我有 5 行代码可以将 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

问题出在这一行： print_r($csvToJson);。 这是输出：

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

这是我的临时文件的文件路径，我做错了什么？

这是我的 CSV 的样子 -

CSV Demo

更新：我的 JSON 格式不正确，名称旁边有 " 和 \

{"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]]"]}

应该怎样——

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

当我打印$csv

Answer 1

看起来您正在将 CSV 的路径输入到 str_getcsv

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

而不是您的实际 CSV 内容

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

Answer 2

您正在将文件分配给变量$csv，但是您正在这样做：

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

变量$csvFile 仍然包含您之前设置为$sourcePath 的文件的文件路径。从file_get_contents() 返回的实际文件字符串在$csv 中。如果你把它改成这样：

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

您可能会发现这可以解决您的问题。

这里是the docs 代表file_get_contents()