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Php 在一段时间内从交易记录中确定一个值_Php_Mysqli - Fatal编程技术网

Php 在一段时间内从交易记录中确定一个值

Php 在一段时间内从交易记录中确定一个值,php,mysqli,Php,Mysqli,我有一个表格可以记录客户的日常交易。但是,如果客户是第一次付款,则无论是在月初、月中还是月底,都应将其标记为1。是否有解决此问题的方法。下面是我的试用版。当它是一个新客户,而不是数据库中已经有交易的现有客户时,它会起作用 <?php if(isset($_POST["save"])){ $custid = $_POST["custid"]; $transDate = $_POST["transDate"]; $grpid = $_POST["cust

我有一个表格可以记录客户的日常交易。但是,如果客户是第一次付款,则无论是在月初、月中还是月底,都应将其标记为1。是否有解决此问题的方法。下面是我的试用版。当它是一个新客户,而不是数据库中已经有交易的现有客户时,它会起作用

    <?php
    if(isset($_POST["save"])){
    $custid = $_POST["custid"];
    $transDate = $_POST["transDate"];
    $grpid = $_POST["custgrp"];
    $amount = $_POST["amount"];
    $postedBy = $_SESSION["staffid"];
    $postStatus= $_POST["postStatus"];
//check if today is first day of the month.if yes, set to 1
    $dF = strtotime($transDate); 
    $rStatus = date("d",$dF);
    if($rStatus==1){
    $rStatus = 1; }else{ $rStatus = 0;} 
//check if the customer has record at all. if yes set the counter to 1 otherwise increase the counter
    $getRole = $connection->query("SELECT * FROM tab_customer_dailycontribution WHERE custid ='$custid'");
    $count = mysqli_num_rows($getRole);

    if($count>=0){
        $count = $count+1;

    $insert = mysqli_query($connection,"INSERT INTO tab_customer_dailycontribution(custid,grpid,amountContribute,transactionDate,counter,rstatus,postedBy,postStatus)VALUES(
              '$custid','$grpid','$amount','$transDate','$count','$rStatus','$postedBy','$postStatus')");

    if($insert){
    echo "<span style='font-weight:bold;color:red;>'"." Transaction is Successful!"."</span>";  

    }else{
    //Unable to save    
    echo "<span style='font-weight:bold;color:black;>'"."Error! Transaction not Successful!"."</span>"; 
    }   
    }//end if count
        //}//end if for checking if today is 1
    }else{
    $custid = "";
    $transDate = "";
    $grpid = "";
    $amount = "";
    $postedBy = ""; 
    }
    ?>

仅在$count>0时使用{@Saty:如果我使用$count>0,对于数据库中没有任何以前记录的新用户会发生什么?而且,我也尝试过,它也不起作用。