php - 提交表单后生成引用ID

每当用户提交表单并将其插入时，我都试图生成像这样的参考ID。

我想出的是在我提交时将CTS-P 0 then CTS-P 1插入数据库。但是问题是在我再次提交后它没有增加CTS-P 0。

我尝试使用CTS-P 0 to CTS-P 1这是到目前为止我所做的。这是最小的事情，但我无法解决。请看一看

    <?php
$con=mysql_connect("localhost","root","");
mysql_select_db("dbname",$con);
$genid="";

if(isset($_POST['submit'])) {
$frstname=$_POST["frstname"];

$genid=mysql_insert_id();
$genid .=count($genid);
//echo $genid;

for($i=0; $i<$genid; $i++) {

$sql = "INSERT INTO tblname (`namecol`,`refidcol`) VALUES ('$frstname','CTS-P $genid[$i]')";
$result = mysql_query($sql);
}}
    ?>

//here is the form
<form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post" name="generate" >

First Name<input type="text"  name="frstname" />
<input type="submit" name="submit" value="Submit" />
</form>

它正在将其作为CTS-P 0插入我的refid列中，但是从下次我提交时开始不递增。

最佳答案

<?php
    $con=mysql_connect("localhost","root","");
    mysql_select_db("dbname",$con);

    if(isset($_POST['submit'])) {
        $frstname=$_POST["frstname"];

        $sql = "SELECT * FROM tblname";
        $genid = mysql_query($sql, $con);
        $genid = mysql_num_rows($genid);

        //Since you're using "0" as your first number, I decided to comment this out, if not, uncomment it
        //$genid++;

        $sql = "INSERT INTO tblname (`namecol`,`refidcol`) VALUES ('$frstname','CTS-P $genid')";
        $result = mysql_query($sql);
    }
?>