mysqli更新不更新

Question

我刚刚将一些旧代码从 mysql 转换为 mysqli。我连接到数据库并读取 AgentNumber 的值，然后该号码将获得分配给的某个电子邮件地址。这一切都很好，电子邮件被发送到不同的电子邮件地址。但是，我想在提交表单后更新数据库，但我无法让更新功能正常工作。 首先是我的数据库的截图： screenshot of database

这里是连接成功后的代码sn-ps： // 将数据库中的表名放入变量中。 $agents = "代理";

    // The query against the database.
    $extractID = "SELECT * FROM $agents";

    // Create a query against the database as a variable as with variables for connection ($mysqli) and for query ($extractID)
    $result = mysqli_query($mysqli, $extractID);

    // Check which number is currently in the database as AgentNumber and use that number to assign the email address
    while($row = mysqli_fetch_assoc($result))
    {
        $agent_number = $row['AgentNumber'];
        // echo $agent_number  ."<br>";
        if ($agent_number == '00056') { 
            $agent_number++; // to 00057
            $insertNew = mysqli_query("UPDATE $agents SET AgentNumber=$agent_number WHERE AgentID=1");
        } elseif ($agent_number == '00057') {
            $agent_number++; // to 00058
            $insertNew = mysqli_query("UPDATE $agents SET AgentNumber=$agent_number WHERE AgentID=1");
        } else {
            $agent_number--;
            $agent_number--; // back to 00056
            $insertNew = mysqli_query("UPDATE $agents SET AgentNumber=$agent_number WHERE AgentID=1");
        }   
    }

我只是无法更改数字，我尝试了很多我在这里找到的陈述。 非常感谢任何帮助。

奥蒂

Answer 1

我忽略了我的号码没有前导零，我还需要再次将连接添加到我的更新查询中。 这是我的工作代码sn-p： // 对数据库的查询。 $extractID = "SELECT * FROM $agents";

    // Create a query against the database as a variable as with variables for connection ($mysqli) and for query ($extractID)
    $result = mysqli_query($mysqli, $extractID);

    // Check which number is currently in the database as AgentNumber and use that number to assign the email address
    while($row = mysqli_fetch_assoc($result))
    {
        $agent_number = $row['AgentNumber'];
        $agent_id = 00001;
        // echo $agent_number  ."<br>";
        if ($agent_number == '00056') { 
            $agent_number = '00057'; // to 00057 - not really necessary to set here as usesd as number in update query
            // echo $agent_number  ."<br>";
            $insertNew = mysqli_query($mysqli, 'UPDATE `agents` SET `AgentNumber` = "00057" WHERE `AgentID`="00001"');
        } elseif ($agent_number == '00057') {
            $agent_number = '00058'; // to 00058 - not really necessary to set here as usesd as number in update query
            $insertNew = mysqli_query($mysqli, 'UPDATE `agents` SET `AgentNumber` = "00058" WHERE `AgentID`="00001"');
        } else {
            $agent_number = '00056'; // to 00056 - not really necessary to set here as usesd as number in update query
            $insertNew = mysqli_query($mysqli, 'UPDATE `agents` SET `AgentNumber` = "00056" WHERE `AgentID`="00001"');
        }   
    }

Answer 2

您的 WHERE 子句可能永远不会命中正确的记录，因为代码尝试设置 WHERE AgentId =1，同时它们在数据库中的 ID 类似于“0001”。

这应该可行：

while($row = mysqli_fetch_assoc($result))
{
    $agent_number = $row['AgentNumber'];
    $agent_id = $row['AgentID'];
    // echo $agent_number  ."<br>";
    if ($agent_number == '00056') { 
        $agent_number++; // to 00057
        $insertNew = mysqli_query($conn, "UPDATE $agents SET AgentNumber=$agent_number WHERE AgentID=$agent_id");

（为清楚起见省略了其余代码）