Question

自我解决（参见底部代码段）

背景知识：我正在为小屋租赁服务建立一个网页。目前我正在显示有关特定小屋的信息，其中使用php脚本从MySql数据库中检索数据：

<?php
$servername = "---------";
$username = "----------";
$password = "----------";
$dbname = "----------";

// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);

// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
} 

$cottage = 1;

$cottage_name = "SELECT cottage FROM AidenLockeRates where id = 'cottage'";

$RegRate = "SELECT RegRate FROM AidenLockeRates where id = 'cottage'";

$PeakRate = "SELECT PeakRate FROM AidenLockeRates where id = 'cottage'";

$Blurb = "SELECT Blurb FROM AidenLockeRates where id = 'cottage'";

echo $cottage "<br>";

echo $cottage_name "<br>";

echo $RegRate "<br>";

echo $PeakRate "<br>";

echo $Blurb "<br>";

?>

问题：在http://phpfiddle.org“编译”脚本时出现错误：

Line : 25,   Error type : 4
Message : syntax error, unexpected '"
"' (T_CONSTANT_ENCAPSED_STRING), expecting ',' or ';'

我的脚本中突出显示与错误相关的行。我已经尝试了几种语法变体，我不知道还能做什么！希望及时回复，谢谢！

感谢评论员的帮助和挖掘其他论坛，我找到了遇到类似问题的人的解决方案：

<?php
$servername = "-------";
$username = "-------";
$password = "-------";
$dbname = "-------";

// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);

// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
}

else {
echo "conn success <br>";

}

$cottage = 1;

$cottage_name = mysqli_query($conn,"SELECT `cottage` FROM `AidenLockeRates` WHERE id = '$cottage'");

$reg_rate = mysqli_query($conn,"SELECT `RegRate` FROM `AidenLockeRates` WHERE id = '$cottage'");

$peak_rate = mysqli_query($conn,"SELECT `PeakRate` FROM `AidenLockeRates` WHERE id = '$cottage'");

$blurb = mysqli_query($conn,"SELECT `Blurb` FROM `AidenLockeRates` WHERE id = '$cottage'");

echo $cottage . "<br>";

while($cottage_name2 = mysqli_fetch_array($cottage_name))
echo $cottage_name2['cottage'] . "<br>";

while($reg_rate2 = mysqli_fetch_array($reg_rate))
echo $reg_rate2['RegRate'] . "<br>";

while($peak_rate2 = mysqli_fetch_array($peak_rate))
echo $peak_rate2['PeakRate'] . "<br>";

while($blurb2 = mysqli_fetch_array($blurb))
echo $blurb2['Blurb'] . "<br>";

$conn->close();

?>

SELECT FROM where语句中的意外''

0 个答案: