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我創建一個函數來找到所有在我的數據庫名字和姓氏的所有我想,如果這些數據是已經存在我只是想輸出錯誤信息功能,如果數據是在數據庫中
我的問題是如何創建一個函數來檢查數據是否已經存在?
這是我的函數來查找名和姓的所有數據。
function find_student_by_firstname($firstname){
global $con;
$safe_firstname = prep($firstname);
$sql = "SELECT * ";
$sql .= "FROM studeprofile ";
$sql .= "WHERE FirstName = '{$safe_firstname}' ";
$sql .= "LIMIT 1";
$student_set = mysqli_query($con, $sql);
confirm_query($student_set);
if($student = mysqli_fetch_assoc($student_set)){
return $student;
} else {
return null;
}
}
function find_student_by_lastname($lastname){
global $con;
$safe_lastname = prep($lastname);
$sql = "SELECT * ";
$sql .= "FROM studeprofile ";
$sql .= "WHERE LastName = '{$safe_lastname}' ";
$sql .= "LIMIT 1";
$student_set = mysqli_query($con, $sql);
confirm_query($student_set);
if($student = mysqli_fetch_assoc($student_set)){
return $student;
} else {
return null;
}
}
這是我目前檢查數據是否已經存在的函數。
function match_fistname_lastname($lastname, $firstname){
$student_firstname = find_student_by_firstname($lastname);
if($student_firstname){
find_student_by_lastname($lastname);
} else {
return false;
}
}
您知道有關SELECT查詢'和'操作?只要選擇那些姓氏與'姓氏'姓氏匹配的條目即可。 – arkascha
謝謝,你的建議有助於解決我的問題。 – jhunlio