Question

有什么理由说这不起作用吗？

<form method="post" action="CorseProject.php">
    <table width="400" border="0" cellspacing="1" cellpadding="2">
        <tr>
            <td width="100">EmployeeID</td>
            <td><input name="employeeid" type="text" id="employeeid" required oninvalid="this.setCustomValidity('You must enter an employee id')" onchange="this.setCustomValidity('')"></td>
        </tr>
        <tr>
            <td width="100">ManagementID</td>
            <td><input name="managementid" type="text" id="managementid" required oninvalid="this.setCustomValidity('You must enter a management id')" onchange="this.setCustomValidity('')"></td>
        </tr>
        <tr>
            <td width="100">First Name</td>
            <td><input name="first_name" type="text" id="first_name" required oninvalid="this.setCustomValidity('You must enter a first name')" onchange="this.setCustomValidity('')"></td>
        </tr>
        <tr>
            <td width="100">Last Name</td>
            <td><input name="last_name" type="text" id="last_name" required oninvalid="this.setCustomValidity('You must enter a last name')" onchange="this.setCustomValidity('')"></td>
        </tr>
        <tr>
            <td width="100">Age</td>
            <td><input name="age" type="text" id="age" required oninvalid="this.setCustomValidity('You must enter an age)" onchange="this.setCustomValidity('')"></td>
        </tr>
        <tr>
            <td width="100">Employment Period in Months</td>
            <td><input name="employment_period" type="text" id="employment_period" required oninvalid="this.setCustomValidity('You must enter an employment period')" onchange="this.setCustomValidity('')"></td>
        </tr>
        <tr>
            <td width="100"> </td>
            <td><input name="find" type="submit" id="add" value="Insert New Record"></td>
        </tr>
    </table>
</form>

下面的我的配置和opendb文件工作正常，因此它们不会导致此错误。

<?php 
    include 'mod2_config.php';
    include 'mod2_opendb.php';
    $employeeid = (isset($_POST['employeeid'])? $_POST['employeeid']:'');
    $managementid = (isset($_POST['managementid']) ? $_POST['managementid']: '');
    $fname= (isset($_POST['first_name'])    ? $_POST['first_name']   : '');
    $lname= (isset($_POST['last_name'])    ? $_POST['last_name']   : '');
    $age= (isset($_POST['age'])    ? $_POST['age']   : '');
    $employment_period= (isset($_POST['employment_period'])    ? $_POST['employment_period']   : ''); 
    $sql= " INSERT INTO employees (employeeid, managementid, first_name,last_name, age, employement_period)
    VALUES ('$employeeid','$managementid',$fname','$lname','$age', '$employment_period')";
    $result = mysqli_query($con, $sql);
    $sql= " SELECT employeeid, managementid, first_name, last_name, age,employment_period from employees WHERE employeeid = $employeeid LIMIT 1";
    $result = mysqli_query($con, $sql);
    if (mysqli_num_rows($result) > 0) {
        // output data of each row
        while($row = mysqli_fetch_assoc($result)) {
            echo "<b>Record successfully inserted:</b><br>";
            echo "<b>EmployeeID: " . $row["employeeid"]. "</b><br>";
            echo "<b>ManagementID: " . $row["managementid"]. "</b><br>";
            echo "<b>Name: " . $row["first_name"]. " " . $row["last_name"]. "</b><br>";
            echo "<b>Age: " . $row["age"]. "</b><br>";
            echo "<b>Employment Period In Months: " . $row["employment_period"]. "</b><br>";
        }
    }
    else {
        echo "Sorry there are no matches! Please check your entry and try again.";
    }

    //The else statement above keeps running instead of the form inserting information into the database and showing the results.
    mysqli_close($con);
?>

我已经尝试了几个小时的插入脚本，但仍然无法将其插入到数据库中。

Answer 1

试试这个。

<?php 
include 'mod2_config.php';
include 'mod2_opendb.php';


$employeeid= (isset($_POST['employeeid'])    ? $_POST['employeeid']   : '');
$managementid= (isset($_POST['managementid'])    ? $_POST['managementid']   : '');
$fname= (isset($_POST['first_name'])    ? $_POST['first_name']   : '');
$lname= (isset($_POST['last_name'])    ? $_POST['last_name']   : '');
$age= (isset($_POST['age'])    ? $_POST['age']   : '');
$employment_period= (isset($_POST['employment_period'])    ? $_POST['employment_period']   : ''); 


$sql_insert = " INSERT INTO employees (employeeid, managementid, first_name, last_name, age, employement_period)
                    VALUES ('$employeeid','$managementid','$fname','$lname','$age', '$employment_period')";
$result_insert = mysqli_query($con, $sql_insert);

$sql_fetch = " SELECT * from employees WHERE employeeid = $employeeid LIMIT 1";

$result_fetch = mysqli_query($con, $sql_fetch);

if (mysqli_num_rows($result_fetch) > 0) {
  // output data of each row
  while($row = mysqli_fetch_assoc($result_fetch)) {
    echo "<b>Record successfully inserted:</b><br>";
    echo "<b>EmployeeID: " . $row["employeeid"]. "</b><br>";
    echo "<b>ManagementID: " . $row["managementid"]. "</b><br>";
    echo "<b>Name: " . $row["first_name"]. " " . $row["last_name"]. "</b><br>";
    echo "<b>Age: " . $row["age"]. "</b><br>";
    echo "<b>Employment Period In Months: " . $row["employment_period"]. "</b><br>";
  }
}   
    else {
    echo "Sorry there are no matches! Please check your entry and try again.";
}

Answer 2

错误位于 $ fname

的插入查询中

将 $ fname'替换为'$ fname'

$sql= " INSERT INTO employees (managementid, first_name,last_name, age, employement_period)
VALUES ('$managementid','$fname','$lname','$age', '$employment_period')";

这样做

制作自动递增的主键。它可能是 employee_id 。
插入时不要在查询中添加该主键，因为它是自动增量。

然后像这样编写选择查询。

$sql= " SELECT employeeid, managementid, first_name, last_name, age,employment_period from employees ORDER BY employeeid DESC LIMIT 1";

它会为您提供所需的结果....

PHP表单不返回数据

2 个答案: