Php 引用完整性错误

我在数据库期末考试项目中遇到了一些引用完整性错误。我自己也试过想办法，但没用，我希望这里的人能给我指明正确的方向

问题是我有三个表，前两个是第三个表的父表。我试图构造一个php脚本，它将更新两个父表，然后将数据插入子表。到目前为止，我认为我所拥有的一切都是可行的，但每当我尝试使用它时，我都会出现以下错误：

在采购中插入CustomerID、PurchaseOrderNo、PurchaseTotal、DateOfPurchase、SalesorSoID、SpecialLorder值“10”、“0000”、“100.00”、“0000-00-00”、“5555”、“N”

无法添加或更新子行：外键约束“xxxxxx”失败。“购买”，约束“购买”\u ibfk\u 1“外键”CustomerID“在更新级联时删除级联时引用“CustomerInfoCustomerID”

你知道我哪里出了问题吗

        ########## FOREIGN KEY CHECK START ##########
$sql = "select count(*) as count from CustomerInfo where '$customerid' = CustomerID";
$result = mysqli_query($con,$sql)
or  die('Error: ' . mysql_error());

$row = mysqli_fetch_assoc($result);
    if ( $row['count']==0 ){
    "INSERT INTO CustomerInfo ('CustomerID') VALUES ('$customerid');";
    echo "<p>Customer ID Not Found. <br />New CustomerID Created.</p>";
   }

$sql2 = "select count(*) as count from EmployeeInfo where '$salespersonid' =  SalesPersonID;";
$result2 = mysqli_query($con,$sql)
 or  die('Error: ' . mysql_error());
 $row2 = mysqli_fetch_assoc($result2);

if ( $row2['count']==0 ){
"INSERT INTO EmployeeInfo ('SalesPersonID')VALUES ('$salespersonid');";
echo "<p>Salesperson ID Not Found. <br />New Salesperson ID Created.</p>";
}

    ########## FOREIGN KEY CHECK END ##########

    ########## DATA ENTRY SQL STATEMENT START ##########
$sql3 = "INSERT INTO Purchases (CustomerID, 
          PurchaseOrderNo, 
          PurchaseTotal, 
          DateOfPurchase, 
          SalesPersonID,
          SpecialOrder) 
VALUES ('$customerid',   
        '$purchaseorderno',
        '$purchasetotal',
        '$dateofpurchase',
        '$salespersonid',
        '$specialorder')";
    ########### DATA ENTRY SQL STATEMENT END ##########

    ########## INPUT SUCCESS/FAILURE REPORTING#########
if (mysqli_query($con, $sql3)) {
    echo "<P>Record Successfully Created</P><BR />";
} else {
  echo "Error: " . $sql9. "<br>" . mysqli_error($con);
}
mysqli_close($con);
echo "<P>Connection Successfully Closed.</P>";

---

您没有对CustomerInfo或EmployeeInfo执行插入查询

你想要：

INSERT INTO CustomerInfo (CustomerID) VALUES ('$customerid')

---

MySQL的确切错误是什么？我从未听说过引用完整性错误…它告诉您，您尝试插入的CustomerID在CustomerInfo表中不存在。如果我们看到表定义，可能会有所帮助。看起来您实际上没有运行针对CustomerInfo的插入查询。。。或为雇员信息

INSERT INTO CustomerInfo ('CustomerID') VALUES ('$customerid')

INSERT INTO CustomerInfo (CustomerID) VALUES ('$customerid')