我遇到这样的错误，idk为什么会出错。
Uncaught mysqli_sql_exception：无法添加或更新子行：外键约束失败（studentsinformation。honor，CONSTRAINT honor_ibfk_1外键（ExamineeID））参考personal（ExamineeID）ON在C：\ xampp \ htdocs \ Practice \ dummy.php：74中删除级联更新堆栈：＃0 C：\ xampp \ htdocs \ Practice \ dummy.php（74）：mysqli_stmt-> execute（）＃1 {main}在第74行的C：\ xampp \ htdocs \ Practice \ dummy.php中抛出
这些是我的代码：
else{
$sql = "SELECT FirstName, MiddleName, LastName, LastSchoolAttended From personal Where FirstName = ? And MiddleName = ? And LastName = ? And LastSchoolAttended = ? Limit 1";
$sqlhonor = "SELECT Honor1, Honor2, Honor3, Honor4 From honor Where Honor1 = ? And Honor2 = ? And Honor3 = ? And Honor4 = ? Limit 0";
$sqlparent = "SELECT Father, EducationalFather, Mother, EducationalMother, Guardian, Occupation From parent Where Father = ? And EducationalFather = ? And Mother = ? And EducationalMother = ? AND Guardian = ? AND Occupation = ? Limit 0";
if ($stmt = $conn->prepare($sql)); {
$INSERTpersonal = "INSERT INTO personal (FirstName, MiddleName, LastName, Age, HomeAddress, ContactNumber, LastSchoolAttended, Strand, SchoolAddress, Adviser) values (?,?,?,?,?,?,?,?,?,?)";
$stmt -> bind_param("ssss", $FirstName,$MiddleName,$LastName,$LastSchoolAttended);
$stmt -> execute();
$stmt -> bind_result($FirstName,$MiddleName,$LastName,$LastSchoolAttended);
$stmt -> store_result();
$rnum = $stmt -> num_rows;
if ($rnum == 0 ) {
$stmt->close();
$stmt = $conn -> prepare($INSERTpersonal);
$stmt -> bind_param("sssisissss", $FirstName,$MiddleName,$LastName,$Age,$HomeAddress,$ContactNumber,$LastSchoolAttended,$Strand, $SchoolAddress,$Adviser);
$stmt -> execute();
}
}
if ($stmthonor = $conn->prepare($sqlhonor)) {
$INSERThonor = "INSERT INTO honor (Honor1,Honor2,Honor3,Honor4) values (?,?,?,?)";
$stmthonor -> bind_param("ssss", $Honor1,$Honor2,$Honor3,$Honor4);
$stmthonor -> execute();
$stmthonor -> bind_result($Honor1,$Honor2,$Honor3,$Honor4);
$stmthonor -> store_result();
$rnum = $stmthonor -> num_rows;
if ($rnum == 0 ) {
$stmthonor->close();
$stmthonor = $conn -> prepare($INSERThonor);
$stmthonor -> bind_param ("ssss", $Honor1, $Honor2, $Honor3, $Honor4);
$stmthonor -> execute();
}
}
if ($stmtparent = $conn -> prepare ($sqlparent)) {
$INSERTparent = "INSERT INTO parent (Father, EducationalFather,Mother,EducationalMother,Guardian,Occupation) values (?,?,?,?,?,?)";
$stmtparent -> bind_param("ssssss", $Father,$EducationalFather,$Mother,$EducationalMother, $Guardian, $Occupation);
$stmtparent -> execute();
$stmtparent -> bind_result($Father,$EducationalFather,$Mother,$EducationalMother,$Guardian ,$Occupation );
$stmtparent -> store_result();
$rnum = $stmtparent -> num_rows;
if ($rnum == 0 ) {
$stmtparent->close();
$stmtparent = $conn -> prepare($INSERTparent);
$stmtparent -> bind_param ("ssssss", $Father, $EducationalFather, $Mother, $EducationalMother, $Guardian, $Occupation);
$stmtparent -> execute();
}
}
}
例如，如果我要填写所需的信息，我希望将它们分发到自己的表中，但仍保持相互连接（这就是我分配FK的原因），但是在我的情况是，他们根本没有分发，只填了个人表，而其他人则是空的，并显示了类似的错误。
我希望数据库成为：enter image description here
解决问题的非常基本且非常糟糕的例子。
NOTE提供的代码只能用作示例，也许可以帮助您解决问题。我在这里尽可能多地发表评论。
首先发布在下面的代码基本上可以确保personal表中存在具有给定参数的记录。这些参数包括ExamineeID。
接下来，它首先尝试使用ExamineeID从honor表中选择数据，并确保先前成功完成ExamineeID的查询。
如果具有此ExamineeID的记录已存在于honor中，那么什么也不会发生，如果该记录不存在，则将其插入其中
该代码正在处理您在This fiddle中提供的数据库结构，>
<?php
mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT); // Turn on mysqli exeptions. Much better than errors.
$conn = new mysqli('localhost', 'user', 'pass', 'db'); //Initialize Db connection
// Generic values for testing
$FirstName = "test";
$MiddleName = "test";
$LastName = "test";
$Age = 1000;
$HomeAddress = "test";
$ContactNumber = 1000;
$LastSchoolAttended = "test";
$Strand = "test";
$SchoolAddress = "test";
$Adviser = "test";
$Honor1 = "test";
$Honor2 = "test";
$Honor3 = "test";
$Honor4 = "test";
$Father = "test";
$EducationalFather = "test";
$Mother = "test";
$EducationalMother = "test";
$Guardian = "test";
$Occupation = "test";
$ExamineeID;
/**
*
* Pay great attention. You have to actually select ExamineeID from table personal, so you need to add it as a selected param.
* Also we limit our result to a single row, ideally this whould not be done with LIMIT 1,
* but in reality we don't have any other way of making sure only one record is selected, using provided.info.
* As I understand it an ExamineeID which solves the selection problem can actually be unknown on the time of selection
*
*/
$sql = "SELECT ExamineeID, FirstName, MiddleName, LastName, LastSchoolAttended From personal Where FirstName = ? And MiddleName = ? And LastName = ? And LastSchoolAttended = ? LIMIT 1";
/**
*
* Pay even greater attention. Honor1, Honor2, Honor3,Honor4 are being selected and should non be in that where clause.
* Determining correct values that should be selected is based on ExamineeID, which should be somehow provided.
* but in reality we don't have any other way of making sure only one record is selected, using provided.info.
* As I understand it an ExamineeID which solves the selection problem can actually be unknown on the time of selection
*
*/
$sqlhonor = "SELECT Honor1, Honor2, Honor3, Honor4 From honor Where ExamineeID = ?";
/**
*
* This statements doesn't relate to problem in any way. Any queries will be omitted as they just bring confusion
*
*/
$sqlparent = "SELECT Father, EducationalFather, Mother, EducationalMother, Guardian, Occupation From parent Where Father = ? And EducationalFather = ? And Mother = ? And EducationalMother = ? AND Guardian = ? AND Occupation = ?";
$stmt = $conn->prepare($sql); // You don't need this if. If this fails then you will get exeption.
$stmt->bind_param("ssss", $FirstName,$MiddleName,$LastName,$LastSchoolAttended);
$stmt->execute();
$stmt->bind_result($ExamineeID,$FirstName,$MiddleName,$LastName,$LastSchoolAttended);
$stmt->store_result(); // As I understand it store_result() doesn't actually fill values specified in bind_result().
$stmt->fetch(); // fetch on the other hand fills bind_result variables.
$stmt->close();
if (!isset($ExamineeID)) { // We don't check the number of rows, we check if and ExamineeID was actually found in previous query
$INSERTpersonal = "INSERT INTO personal (FirstName, MiddleName, LastName, Age, HomeAddress, ContactNumber, LastSchoolAttended, Strand, SchoolAddress, Adviser) values (?,?,?,?,?,?,?,?,?,?)";
$stmt = $conn->prepare($INSERTpersonal);
$stmt->bind_param("sssisissss", $FirstName,$MiddleName,$LastName,$Age,$HomeAddress,$ContactNumber,$LastSchoolAttended,$Strand, $SchoolAddress,$Adviser);
$stmt->execute();
$stmt->close();
// After we add a value to personal table we have to re-select all values.
$stmt = $conn->prepare($sql);
$stmt->bind_param("ssss", $FirstName,$MiddleName,$LastName,$LastSchoolAttended);
$stmt->execute();
$stmt->bind_result($ExamineeID,$FirstName,$MiddleName,$LastName,$LastSchoolAttended);
$stmt->store_result();
$stmt->fetch();
$stmt->close();
}
//Here
if (isset($ExamineeID)) { // We only need to check that we actually found a $ExamineeID earlier.
$stmthonor = $conn->prepare($sqlhonor);
$stmthonor->bind_param("s",$ExamineeID); // We need to bind only a single value as I pointed out earlier.
$stmthonor->execute();
$stmthonor->bind_result($Honor1,$Honor2,$Honor3,$Honor4);
$stmthonor->store_result();
$stmthonor->fetch(); // Again fetch is needed to actually fill Honor vars
$rnum = $stmthonor->num_rows;
$stmthonor->close();
// If no entries that have ExamineeID we found earlier exist in Database.
if ($rnum == 0 ) {
$INSERThonor = "INSERT INTO honor (Honor1,Honor2,Honor3,Honor4,ExamineeID) values (?, ?, ?, ?, ?)"; // We need to insert 5 values!!
$stmthonor = $conn->prepare($INSERThonor);
// We actually insert a record with provided ExamineeID, because earlier there was no such record.
$stmthonor->bind_param ("ssssi", $Honor1, $Honor2, $Honor3, $Honor4, $ExamineeID);
$stmthonor->execute();
$stmthonor->close();
}
}