Question

尝试运行此脚本时，我的浏览器出现以下错误：

 Parse error: syntax error, unexpected end of file in C:\wamp\www\server2.php on line 35

注意：第35行是脚本中的最后一行。对不起，我是PHP的新手

  <?php
      // get the command
      $command = $_REQUEST['command'];

      // determine which command will be run
      if($command == "getAnimalList") {
        // return a list of animals
        echo "bird,dog,cat,cow,sheep";
      } else if($command == "getAnimalSound") {
        // get the animal parameter and send the right response
    $animal = $_REQUEST['cat'];

    // fetch the sound of the animal from the database
    $username = "root"; $password = ""; $hostname = "localhost"; 

    //connection to the database
    $dbhandle = mysql_connect($hostname, $username, $password) 
      or die("Unable to connect to MySQL");
    echo "Connected to MySQL<br>";


    //select a database to work with
    $selected = mysql_select_db("rosslocalhost1",$dbhandle) 
      or die("Could not select examples");

    //execute the SQL query and try to return a record
    $result = mysql_query("SELECT sound FROM animalsounds WHERE name='$animal'");

    if (!$result) {
        echo 'Dont know' .mysql_error();
        exit;
    }
    $row = mysql_fetch_array($result);
    echo $row['sound'];
    ?>

Answer 1

您已错过elseif <{1}}声明中的结束括号

if($command == "getAnimalList")

Answer 2

这是你的代码正确缩进。您没有正确关闭else if ($command == "getAnimalSound")

<?php
// get the command
$command = $_REQUEST['command'];

// determine which command will be run
if ($command == "getAnimalList") {
    // return a list of animals
    echo "bird,dog,cat,cow,sheep";
} else if ($command == "getAnimalSound") {
    // get the animal parameter and send the right response
    $animal = $_REQUEST['cat'];
}

// fetch the sound of the animal from the database
$username = "root";
$password = "";
$hostname = "localhost";

//connection to the database
$dbhandle = mysql_connect($hostname, $username, $password) or die("Unable to connect to MySQL");
echo "Connected to MySQL<br>";


//select a database to work with
$selected = mysql_select_db("rosslocalhost1", $dbhandle) or die("Could not select examples");

//execute the SQL query and try to return a record
$result = mysql_query("SELECT sound FROM animalsounds WHERE name='$animal'");

if (!$result) {
    echo 'Dont know' . mysql_error();
    exit;
}
$row = mysql_fetch_array($result);
echo $row['sound'];
?>

PHP脚本：语法错误，意外的文件结束

2 个答案: