我使用MacBook Pro,我将最新版本的Mojave用于本地主机,我使用XAMPP服务器,当我想与DB连接时,消息就会出现:
# Define a Create() method for the label *type* that acts as a factory
# method and outputs a new label *instance*.
# Note that the method becomes an *instance method* of the *type* object.
Add-Member -Force -InputObject ([System.Windows.Forms.Label]) -MemberType ScriptMethod `
-Name Create -Value { #`#
param(
[int[]] $Location,
[int[]] $Size,
[string] $Text
)
New-Object $this.FullName -Property @{
Location = New-Object Drawing.Point $Location[0], $Location[1]
Size = New-Object Drawing.Size $Size[0], $Size[1]
Text = $Text
}
}
# Call the type's new instance method with the desired arguments.
$amuletLabelValueDmg = [System.Windows.Forms.Label].Create((10,20), (60,20), 'Value')
有时我将端口更改为8080,并且显示了相同的质量,我下载了mysql-connector,但无法解决问题!
答案 0 :(得分:0)
通信链接失败。请根据您的的版本更新JDBC驱动程序。