此代码用于更新客户的访问次数。 因此，当员工按下＆＃34;添加访问＆＃34;按钮，访问次数应增加1，日期也应该是最新的。
代码中没有错误，但我在更新中遇到问题。 问题是访问次数增加，即使用户没有点击＆＃34;添加访问＆＃34;按钮。
所以，我不知道代码中的错误是什么。
那么，你能帮我吗？
//update visit
$visitno= $row['visit'];
$currentdate = date('Y-m-d');
$disc0 = "0%";
$disc1 = "5%";
$disc2 = "10%";
$disc3 = "Free Service";
switch ($visitno){
case "1":
{
echo "<center><b>";
echo "The client has 1 visit";
echo "<br/>";
echo "To update the visits, press the button";
echo "<br/>";
echo "<button id='shiny' onclick='alertFunction()'>Add visit</button>";
echo "</center></b>";
$query1= "UPDATE clients.clients SET visit='$visitno'+1, vdate='$currentdate', dis='$disc0' WHERE $category = '$criteria' AND visit='1'" ;
$result1= mysqli_query($con, $query1);
if(!$result1){
echo "Error ". mysqli_error($con);
echo "<br>";
}//END IF
}// end of case1
break;
case "2":
{
echo "<center><b>";
echo "The client has 2 visit";
echo "<br/>";
echo "To update the visits, press the button";
echo "<br/>";
echo "<button id='shiny' onclick='alertFunction()'>Add visit</button>";
echo "</center></b>";
$query1= "UPDATE clients.clients SET visit='$visitno'+1, vdate='$currentdate', dis='$disc0' WHERE $category = '$criteria' AND visit='2'" ;
$result1= mysqli_query($con, $query1);
if(!$result1){
echo "Error ". mysqli_error($con);
echo "<br>";
}//END IF
}// end of case2
break;
case "3":
{
echo "<center><b>";
echo "The client has 3 visit";
echo "<br/>";
echo "To update the visits, press the button";
echo "<br/>";
echo "<button id='shiny' onclick='alertFunction()'>Add visit</button>";
echo "</center></b>";
$query1= "UPDATE clients.clients SET visit='$visitno'+1, vdate='$currentdate', dis='$disc1' WHERE $category = '$criteria' AND visit='3'" ;
$result1= mysqli_query($con, $query1);
if(!$result1){
echo "Error ". mysqli_error($con);
echo "<br>";
}//END IF
}// end of case3
break;
case "4":
{
echo "<center><b>";
echo "The client has 4 visit";
echo "<br/>";
echo "To update the visits, press the button";
echo "<br/>";
echo "<button id='shiny' onclick='alertFunction()'>Add visit</button>";
echo "</center></b>";
$query1= "UPDATE clients.clients SET visit='$visitno'+1, vdate='$currentdate', dis='$disc2' WHERE $category = '$criteria' AND visit='4'" ;
$result1= mysqli_query($con, $query1);
if(!$result1){
echo "Error ". mysqli_error($con);
echo "<br>";
}//END IF
}// end of case4
break;
case "5":
{
echo "<center><b>";
echo "The client has 5 visit";
echo "<br/>";
echo "To update the visits, press the button";
echo "<br/>";
echo "<button id='shiny' onclick='alertFunction()'>Add visit</button>";
echo "</center></b>";
$query1= "UPDATE clients.clients SET visit='$visitno'+1, vdate='$currentdate', dis= '$disc3' WHERE $category = '$criteria' AND visit='5'" ;
$result1= mysqli_query($con, $query1);
if(!$result1){
echo "Error ". mysqli_error($con);
echo "<br>";
}//END IF
}// end of case5
break;
case "6":
{
echo "<center><b>";
echo "Cannot add more visits!";
echo "</center></b>";
}// end of case6
break;
} //end of switch
echo "<script>
function alertFunction()
{
var visitnu = $visitno +1 ;
alert('Client has ' + visitnu + ' visits now');
}
</script>";
}
?>
行。这是我在编辑更新部分的代码后得到的 但它没有执行该功能。 我不知道我应该将ID放在代码中的特定位置或问题出在哪里。
<?php
$visitno= $row['visit'];
$disc0 = "0%";
$disc1 = "5%";
$disc2 = "10%";
$disc3 = "Free Service";
if ($visitno == 6){
echo "Cannot add more visit!";
updatevisit(6,$disc3);
} else if($visitno == 5){
updatevisit(5,$disc2);
} else if($visitno == 4){
updatevisit(4, $disc1);
} else if($visitno == 3){
updatevisit(3, $disc0);
} else if($visitno == 2){
updatevisit(2, $disc0);
} else if($visitno == 1){
updatevisit(1, $disc0);
}
}
function updatevisit($visitno, $disc) {
$currentdate = date('Y-m-d');
$newvisit = $visitno +1;
$discount = $disc;
echo "The customer has $visitno visits";
echo "Please press the button to update the visits";
echo "<form method='post' action='tryupdate.php'>";
echo "<input type='submit' id='updatevisit' name='updatevisit' value='Add visit'/>";
echo "</form>";
if(isset($_POST['updatevisit'])){
$query1 = "UPDATE clients.clients SET visit='$newvisit', vdate='$currentdate', dis='$discount'" ;
$result1 = mysqli_query($con, $query1) or die ('Error Updating');
}
}
?>
那么，你能告诉我哪里出错了。
答案 0 :(得分：0)
在您编辑代码之前，我完成了代码，这对我来说很重要，可能与您的问题有关：
// this line is not finished
$result= mysqli_query($con, $quer
其他一些说明：
// align is mispelled
echo '<tr style="background-color: #80E6CC;color:white;" aligh="center">';
// incorrect use of styles, should be: style='padding:15px'
echo "<tr><td align='center'padding:15px>";
// you don't need brackets here
case "1":
{
如果刷新页面或按下后退按钮，则没有任何内容可以阻止重新提交表单。
进行一些基本的错误检查。逐行遵循代码的逻辑流程并输出：
echo 'here';
exit;
直到您缩小具体问题区域。打开错误报告和/或检查日志。