将数据写入表时出现语法错误 - 意外的T_VARIABLE

时间:2014-02-04 01:38:25

标签: php insert syntax-error

解析错误:语法错误,第44行的upload_file.php中出现意外的T_VARIABLE

代码一直有效,直到我添加这些行:

第42-44行:

$path = "uploads/" . $_FILES["file"]["name"];
$Link = mysql_connect($Host, $User, $Password);
$Query = "INSERT INTO $Table_7 VALUES ('0','"$path"')";

非常感谢。该脚本用于将图像上载到文件夹中。那部分工作,但我无法将图像路径写入表中。我有一个包含两个字段的表:

picid - auto incrementing primary key
path - varchar(60)

知道我做错了什么吗?我添加了完整的脚本。

更新。完整代码

<?php
include "connect.php";
$allowedExts = array("gif", "jpeg", "jpg", "png");
$temp = explode(".", $_FILES["file"]["name"]);
$extension = end($temp);
if ((($_FILES["file"]["type"] == "image/gif")
|| ($_FILES["file"]["type"] == "image/jpeg")
|| ($_FILES["file"]["type"] == "image/jpg")
|| ($_FILES["file"]["type"] == "image/pjpeg")
|| ($_FILES["file"]["type"] == "image/x-png")
|| ($_FILES["file"]["type"] == "image/png"))
&& ($_FILES["file"]["size"] < 10000)
&& in_array($extension, $allowedExts))
  {
  if ($_FILES["file"]["error"] > 0)
    {
    echo "Return Code: " . $_FILES["file"]["error"] . "<br>";
    }
  else
    {
    echo "Upload: " . $_FILES["file"]["name"] . "<br>";
    echo "Type: " . $_FILES["file"]["type"] . "<br>";
    echo "Size: " . ($_FILES["file"]["size"] / 1024) . " kB<br>";
    echo "Temp file: " . $_FILES["file"]["tmp_name"] . "<br>";

    if (file_exists("uploads/" . $_FILES["file"]["name"]))
      {
      echo $_FILES["file"]["name"] . " already exists. ";
      }
    else
      {
      move_uploaded_file($_FILES["file"]["tmp_name"],
      "uploads/" . $_FILES["file"]["name"]);
      echo "Stored in: " . "uploads/" . $_FILES["file"]["name"];
      }
    }
  }
else
  {
  echo "Invalid file";
  }
$path = "uploads/" . $_FILES["file"]["name"];
$Link = mysql_connect($Host, $User, $Password);
$Query = "INSERT INTO $Table_7 VALUES ('0','{$path}')";
?>

1 个答案:

答案 0 :(得分:2)

第44行你错过了连接运算符:

$Query = "INSERT INTO $Table_7 VALUES ('0','"$path"')";

应该是

$Query = "INSERT INTO $Table_7 VALUES ('0','".$path."')";

$Query = "INSERT INTO $Table_7 VALUES ('0','$path')";

$Query = "INSERT INTO $Table_7 VALUES ('0','{$path}')";