[英]How do get the following PHP echo to output this JavaScript alert with a PHP variable?
我正在使用PHP 5.6使用数据库。
$connect = new mysqli($serverName, $username, $password);
if ($connect->connect_error){
$connectError = mysqli_connect_error();
echo('<script> alert("Connection Failed :/'.$connectError.'")</script>');
}else{
echo(('<script> alert("Connection success :)")</script>'));
}
问题是$ connectError停止输出javaScript警报的代码,我如何包含错误消息并输出警报?
您需要转义$connectError
:
$connect = new mysqli($serverName, $username, $password);
if ($connect->connect_error){
$connectError = addslashes(mysqli_connect_error());
echo('<script> alert("Connection Failed :/'.$connectError.'")</script>');
}else{
echo(('<script> alert("Connection success :)")</script>'));
}
我认为问题在于转义引号。 在我的PHP代码中,这很完美:
$message = "Connection Failed :/" .$connectError;
echo "<script>";
echo "alert(\"" .$message. "\");";
echo "</script>";
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.