include_once "connect_to_mysql.php";
$query = mysql_query("SELECT * FROM data WHERE Email = '$email'");
$num = array();
//Caculate the number of rows that have macthing username
$numrows = mysql_num_rows($query);
$num[] = $numrows;
if ($num != 0){
$data = array('success' => 'true', 'message' => 'Incorrect Email');
echo json_encode($data);
}
不处理if语句,也不回显JSON数据,而是冻结脚本 我做错了什么?
答案 0 :(得分:1)
$result = mysql_query("SELECT COUNT(*) AS num FROM data WHERE Email = '$email'");
$row = mysql_fetch_array($result, MYSQL_ASSOC);
$num = intval($row['num']);
if ($num != 0) {
$data = array('success' => 'true', 'message' => 'Email found');
} else {
$data = array('success' => 'false', 'message' => 'Incorrect Email');
}
echo json_encode($data);
答案 1 :(得分:1)
我不明白你为什么要使用$num[]
。
include_once "connect_to_mysql.php";
$email = $_POST['email'];
$query = mysql_query("SELECT * FROM data WHERE Email = '$email'");
$numrows = mysql_num_rows($query);
if ($numrows != 0){
$data = array('success' => 'true', 'message' => 'Correct Email');
echo json_encode($data);
}else{
$data = array('success' => 'false', 'message' => 'Incorrect Email');
echo json_encode($data);
}
尝试使用以下Jquery:
$.post('your_file.php',{email: 'your_email'},function(data){
alert(data.success + ' ' + data.message);
},'json');
答案 2 :(得分:0)
您正在向数组$numrows
中提取标量值$num[]
,因此您遇到了错误。
尝试直接在if语句中使用$numrows
,如下所示:
if ($numrows != 0){
答案 3 :(得分:0)
$query = mysql_query("SELECT * FROM data WHERE Email = '$email'");
if(mysql_num_rows($query)){
$data = array('success' => 'true', 'message' => 'Email found');
}
else{
$data = array('success' => 'false', 'message' => 'Incorrect Email');
}
echo json_encode($data);