Question

你好我的mysql代码有问题，它在尝试输出时出现以下错误：

Parse error: syntax error, unexpected $end in     C:\xampplite\htdocs\learncent\acksearch\search.php on line 25

<?php
$db = new mysqli("localhost","root","","acksocial");

if(mysqli_connect_error())
{
printf("Connection failed:%s \n",mysqli_connect_error());
exit();
}

$name = mysqli_real_escape_string($db, $_POST['search']);
$table = 'acksearch';

if($result = $db->query("SELECT * FROM $table WHERE name = $name", MYSQLI_ASSOC))
{
    while($row = $result->fetch_object())
    {

    // $row is an associative array

    // Do something here

echo "Name: ".$row['name'];

echo " country: ".$row['country'];


}

$db->close();
?>

编辑：添加了1个}并且没有错误但它不会输出结果？

如果有人能帮助我会好的。

再次编辑：现在代码看起来像这样

<?php
$db = new mysqli("localhost","root","","acksocial");

if(mysqli_connect_error())
{
printf("Connection failed:%s \n",mysqli_connect_error());
exit();
}

$name = mysqli_real_escape_string($db, $_POST['search']);
$table = 'acksearch';

if($result = $db->query("SELECT * FROM $table WHERE name = $name", MYSQLI_ASSOC))
{
while($row = $result->fetch_object())
{

// $row is an associative array

// Do something here

echo "Name: ".$row['name'];

echo " country: ".$row['country'];

}

它给出的错误：

Parse error: syntax error, unexpected $end in C:\xampplite\htdocs\searcher.php on line 25

感谢Fredrik

Answer 1

你没有关闭while循环：

while($row = $result->fetch_object())
    {

    // $row is an associative array

    // Do something here

echo "Name: ".$row['name'];

echo " country: ".$row['country'];

}///// I added this } here

您需要在最后一行之后添加}。

Answer 2

如果缺少}某处，则会发生此错误。

在您的情况下，}

之前应该有第二个$db->close();

Answer 3

在结束“？＆gt;”

之前，您需要再关闭一个大括号“}”

输出数据库查询的结果？

3 个答案: