Question

此函数从用户获取日期：

function getCCGraduationDate () {
        ?>
        <form method="post" action="processor.php">
        Graduation Date: <input type="text" name="CCgraduationdate"><br>
        <input type="submit">
        <?php
    }

此函数将值发送到DB，并在processor.php页面上调用：

function sendCCGraduationDate () {
    $con = mysql_connect("localhost","root","XXXXXX");
    if (!$con)
        {
    die('Could not connect: ' . mysql_error());
        }
        mysql_select_db("user", $con);
        $grad = mysql_real_escape_string($_POST['CCGraduationDate'], $con);
        $sql = "UPDATE profile SET CommunityCollegeGraduationDate='$grad' WHERE userid=$this->user_id";
        mysql_query( $sql , $con );

        mysql_close($con);

    }

由于某种原因，该值未在数据库中填充。我昨晚测试了类似的功能，他们工作，不知道我在这里缺少什么。

Answer 1

假设$this->user_id有一个实际值：

$sql = "UPDATE profile SET CommunityCollegeGraduationDate='$grad' WHERE userid=$this->user_id";

// Should be:

$sql = "UPDATE profile SET CommunityCollegeGraduationDate='$grad' WHERE userid={$this->user_id}";

// Or

$sql = "UPDATE profile SET CommunityCollegeGraduationDate='$grad' WHERE userid= " . $this->user_id;

- 编辑 -

$sql = "UPDATE profile SET CommunityCollegeGraduationDate='$grad' WHERE userid= " . $this->user_id;

if (!mysql_query( $sql , $con ))
    die(mysql_error()); 
}

- 编辑 -

// To avoid case and extra space issues, sanitize your var first
$grad = trim(strtolower($grad));

// You did well to escape
$grad = mysql_real_escape_string($grad, $con);

Answer 2

请检查：

表格代码：

<input type="text" name="CCgraduationdate"><br>

PHP代码：

$grad = mysql_real_escape_string($_POST['CCGraduationDate'], $con);

您的表单中的CCgraduationdate在您的php代码中不等于CCGraduationDate

如果您没有收到如下错误：

Notice:  Undefined index: CCGraduationDate /* blah blah blah */

然后您可以尝试在PHP代码的顶部添加：

error_reporting(E_ALL);

然后看看是否出现错误。

当然，这是在将$sql变量的数据更改为：

之后

$sql = "UPDATE profile SET CommunityCollegeGraduationDate='$grad' WHERE userid=" . $this->user_id;

发送到DB功能不执行

2 个答案: