Question

这可能是迄今为止试图获得一些帮助的这篇文章的第四次编辑。我已经多次重写了这段代码，但似乎没有什么工作正常。我已经在这个项目上工作了大约5个月，仍然找不到合适的答案。

目前，除了从表单插入数据外，我已经能够使用语法。插入的内容是空值而不是输入字段中的字符串。

最近的迭代是：

$con = mysqli_connect(host, username, password, db_name1);
if($con->connect_errno){
    printf("Connect failed: %s\n", $connection->error);
    exit();
}

$query = "Insert into hdd_tracking (`ticket`,`serial`,`vol_name`,`drivetype`,`project`,`notes`,`encrypt_date`,`encryption_password`,`fedX_tracking`,`fedX_rt_tracking`)
 VALUES (?,?,?,?,?,?,?,?,?,?)";
$stmt = $con->stmt_init();

if($stmt-> prepare($query)){

    $stmt->bind_param('isssssssss', 
    $_POST['ticket'],
    $_POST['serial'],
    $_POST['vol_name'],
    $_POST['drivetype'],
    $_POST['project'],
    $_POST['notes'],
    $_POST['encrypt_date'],
    $_POST['encryption_password'],
    $_POST['fedX_tracking'],
    $_POST['fedX_rt_tracking']);
}

$ticket = ISSET( $_POST['ticket']);
$serial = ISSET( $_POST['serial']);
$vol_name = ISSET( $_POST['vol_name']);
$drivetype = ISSET( $_POST['drivetype']);
$project = ISSET( $_POST['project']);
$notes = ISSET( $_POST['notes']);
$encrypt_date = ISSET( $_POST['encrypt_date']);
$encryption_password = ISSET( $_POST['encryption_password']);
$fedX_tracking = ISSET( $_POST['fedX_tracking']);
$fedX_rt_tracking = ISSET( $_POST['fedX_rt_tracking']);

$run = $stmt ->execute();
var_dump($stmt);

if($run){
    $stmt -> close();
    mysqli_close($con);
    }else{
        echo "Failure";
    }

表格如下：

<form class="form-inline insertDrive" role="form" name="insert" method="post" action="<?php echo $_SERVER['PHP_SELF']; ?>">
<table class="insertHDD" style="border-collapse: separate; border-spacing: 10px;">
    <tr align="left">
        <td>
            <label class="center">Ticket:</label>
        </td>
        <td>
            <input type="email" class="form-control" name="ticket" placeholder="Ticket">
        </td>
    </tr>
    <tr>
        <td>
            <label class="center">Volume Name:</label>
        </td>
        <td>
            <input type="text" class="form-control" name="vol_name" placeholder="Volume Name">
        </td>
        <td>
            <label class="center">Type:</label>
        </td>
        <td>
            <input type="email" class="form-control" name="drivetype" placeholder="Type">
        </td>
    </tr>

    <tr>
        <td>
            <label class="center">Project:</label>
        </td>
        <td>
            <input type="text" class="form-control" name="project" placeholder="Project">
        </td>
        <td>
            <label class="center">Encryption:</label>
        </td>
        <td>
            <input type="email" class="form-control" name="encryption_password" placeholder="Encryption">
        </td>
    </tr>

    <tr>
        <td>
            <label class="center">Date:</label>
        </td>
        <td>
            <input type="text" class="form-control" name="encrypt_date" placeholder="Date">
        </td>
        <td>
            <label>Serial:</label>
        </td>
        <td>
            <input type="email" class="form-control" name="serial" placeholder="Serial">
        </td>
    </tr>

    <tr>
        <td>
            <label class="center">Tracking:</label>
        </td>
        <td>
            <input type="text" class="form-control" name="fedX_tracking" placeholder="Tracking">
        </td>
        <td>
            <label class="center">Return:</label>
        </td>
        <td>
            <input type="text" class="form-control" name="fedX_rt_tracking" placeholder="Return">
        </td>
    </tr>
    <tr>
        <td>
            <label class="center">Notes:</label>
        </td>
        <td colspan="3">
            <input type="text" width="200px" class="form-control" name="notes" placeholder="Notes">
        </td>
    </tr>
    <tr>
        <td colspan="4">
            <a class="btn btn-small btn-info" name="submit" name="submit">Submit</a>
        </td>
    </tr>
</table>
</form>

我已经使用id到name进行了更正，我仍然得到空值。 var_dump($stmt)返回： object（mysqli_stmt）＃1（10）{[＆＃34; affected_rows＆＃34;] =＆gt; int（1）[＆＃34; insert_id＆＃34;] =＆gt; int（1262）[＆＃34; num_rows＆＃34;] =＆gt; int（0）[＆＃34; param_count＆＃34;] =＆gt; int（10）[＆＃34; field_count＆＃34;] =＆gt; int（0）[＆＃34; errno＆＃34;] =＆gt; int（0）[＆＃34;错误＆＃34;] =＆gt; string（0）＆＃34;＆＃34; [＆＃34; error_list＆＃34;] =＆GT; array（0）{} [＆＃34; sqlstate＆＃34;] =＆gt; string（5）＆＃34; 00000＆＃34; [＆＃34; ID＆＃34;] =＆GT; int（1）}

仍然没有得到这个。

Answer 1

是问题是您正在使用id，您应该在html表单name中使用input。例如

<input type="email" class="form-control" id="ticket" placeholder="Ticket">

应该是

<input type="email" class="form-control" name="ticket" placeholder="Ticket">

使用php $_POST['']时，您正在按名称

访问输入

我试图弄清楚如何使用php / mysqli将数据插入我的数据库

1 个答案: