我最近实现了许多if
语句,用于检查是否已输入需求数据,如果没有输入,则会收到错误消息something is wrong with...
。但是在实现它们之后,无论数据是否实际上已经发送到数据库(正在发送的数据都是正确的),我现在都可以收到该错误消息,而且我一生都无法弄清楚原因。>
$query = "insert into $sql_table (Eoi, Job_reference_number, First_Name, Last_Name, Street_Address, Suburb, State, Postcode, Email, Phone_Number, Skills) values ('$eoi','$jobNumber', '$firstName', '$lastName', '$streetAddress', '$suburb', '$state', '$postcode', '$emailAddress', '$phoneNumber', '$skills')";
$result = mysqli_query($conn, $query);
if($jobNumber = ''){
$result = false;
}
if($firstName = ''){
$result = false;
echo "<p> Something is wrong with your First Name </p>";
}
if($lastName = ''){
$result = false;
echo "<p> Something is wrong with your Last Name </p>";
}
if($streetAddress = ''){
$result = false;
echo "<p> Something is wrong with your Street Address </p>";
}
if($suburb = ''){
$result = false;
echo "<p> Something is wrong with your Suburb </p>";
}
if($postcode = ''){
$result = false;
echo "<p> Something is wrong with your Postcode </p>";
}
if($email = ''){
$result = false;
echo "<p> Something is wrong with your Email </p>";
}
if($phoneNumber = ''){
$result = false;
echo "<p> Something is wrong with your Phone Number </p>";
}
if($skills = ''){
$result = false;
echo "<p> Something is wrong with your Skills </p>";
}
if($result != mysqli_query($conn, $query)) {
echo "<p>Something is wrong with ", $query, "</p>";
}else {
echo "<p class=\"ok\">Successfully added a New EOI record</p>";
}
}
}
mysqli_close($conn);
我希望当用户输入有效数据时结果为Successfully added a new EOI record
,但我收到错误消息。
答案 0 :(得分:0)
首先,if语句中出现语法错误 if语句应为
==
而不是=
if($yourVariable == ''){
echo "<p> Something is wrong with your yourVariable </p>";//no meaning of this line
$result = false;
}
这意味着,如果您的variable
是empty
,那么$result
将false
,您可以在最后一个条件中查看它
第二步,您正在检查数据库插入后的所有变量,需要在插入数据库之前进行检查