我需要一些帮助,在我的表单中创建一个自动递增注册ID。它可能是数字或字母数字。如果它像aaa0001
一样开始我在表单中显示它。在第一个条目之后,它应该增加到aaa0002
并在表单中显示它。我试过JSON和PHP,有些错误。那可能是aaa0001,aaa0002,...... aaa9999 ...... aab0000 ......
PHP
$details = mysqli_query($con, "SELECT MAX(`member_id`) FROM `member_details`");
$details = mysqli_query($con, "SELECT * FROM `member_details` WHERE `member_id` ");
$rows = array();
while ($r = mysqli_fetch_assoc($details)) {
$rows[] = $r;
}
print json_encode($details);
JSON
var xmlhttp = new XMLHttpRequest();
xmlhttp.onreadystatechange = function() {
if (this.readyState == 4 && this.status == 200) {
var myObj = JSON.parse(this.responseText);
//var memid = myObj[0].member_id;
//memid++;
//document.getElementById("member_id").value = memid;
document.getElementById("member_id").value = myObj[0];
alert(myObj);
}
};
xmlhttp.open("GET", "/fetch_reg_memid.php?member_id=" + member_id, true);
xmlhttp.send();
}
首先,您的第一个查询是可以的,有一个更简单的选择:
$details= mysqli_query($con, "SELECT MAX(`member_id`) FROM `member_details`");
$details = mysqli_query($con, "SELECT member_id FROM `member_details` ORDER BY member_id dESC");
第二,应该修复mysqli的提取。 details是json编码的对象
如果只有一行,则不需要遍历案例......所以
$r = mysqli_fetch_assoc($details);
然后$r['member_id']
应该返回你的member_id:
$r = mysqli_fetch_assoc($details);
print $r['member_id'];