问题描述
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我正在尝试使用jquery post函数返回错误以检索错误。
到目前为止,我已经用PHP编写了
$connect = MysqL_connect(\"localhost\",\"username\",\"password\") or die(\"Error connecting to the database.\");
但我想像这样重新发送消息
$connect = MysqL_connect(\"localhost\",\"password\");
if(!$connect){
$data[\'error\'] = true;
$data[\'message\'] = \"Error connecting to the database.\";
echo json_encode($data);
};
救命!!
解决方法
尝试这个:
$connect = mysql_connect(\"localhost\",\"username\",\"password\") or die(json_encode(array(\'error\' => true,\'message\' => mysql_error())));
这会将错误编码为json字符串