动态表单无法正确插入到 MySQL 数据库中

Dynamic Forms Won't Correctly Insert Into MySQL Database

本文关键字：MySQL 数据库 插入 表单 动态      更新时间：2023-09-26
<body>
     <?php
        $con = mysqli_connect('localhost','root','','cash');
        $query = "SELECT DISTINCT category FROM cash";
        $result = mysqli_query($con,$query);
        $dropDownList = '<select name="names[]"><option value = "">---Select---</option>';
        while ( $d=mysqli_fetch_assoc($result)) {
            $dropDownList .= "<option value='" . $d['category'] . "'>" . $d['category'] . "</option>";
        }
        $dropDownList .= '</select>';
    ?>     

    <script type="text/javascript">
    $(document).ready(function() {
        var InputsWrapper = $("#InputsWrapper");
        var AddButton = $("#AddMoreFileBox");
        var dropOption = <?php echo json_encode($dropDownList) ?>;
        var x = InputsWrapper.length;
        var FieldCount = 1;
        $(AddButton).click(function(e)//on add input button click
        {
            FieldCount++;
            $(InputsWrapper).append('<tr><td>'+dropOption+'<td><input type="text" name="cate[]" id="categ"/></td><td><input type="number" name="money[]" id="amount"/></td></tr>'); 
            x++;
            return false;
        });
    });
</script>
     <form action="selectxpprocess.php" method="post">
     <table id="InputsWrapper" >
            <tr>
                <span class="small"><a href="#" id="AddMoreFileBox" class="btn btn-info">Add More Field</a></span>
            </tr>
            <tr>
                <td><label for='names[]'>Category:</label></td>
                <td><label for='cate[]'>New Category:</label></td>
                <td><label for='money[]'>Amount:</label></td>
            </tr>
            <tr>
                <td><?php echo $dropDownList?></td>
                <td><input type="text" name="cate[]" id="categ"/></td>
                <td><input type="number" name="money[]" id="amount"/></td>
            </tr>
        </table>
        <input type="submit" />
           </form>
</body>

这是我的第一页。我有一个按钮，当您单击它时，将弹出另一个下拉列表，文本框和数字输入。我想要的条件是，如果在下拉列表中未选择任何内容，则从文本框中获取数据。之后，将相应的金额值传递给数据库。

<?php
 $con = mysqli_connect('localhost','root','','cash');

if($_POST['names'] != '' && $_POST['cate'] == '') {
    foreach($_POST['names'] as $catego) {
        foreach($_POST['money'] as $amo){
            mysqli_query($con,"INSERT INTO cash (category, amount) VALUES ('".$catego."','".$amo."')");
        }
    } 
}else {
    foreach($_POST['cate'] as $categ) {
        foreach($_POST['money'] as $amo){
            mysqli_query($con,"INSERT INTO cash (category, amount) VALUES ('".$categ."','".$amo."')");
        }
    }
}
$_POST=array();
mysqli_close($con);
header("Location: selectxp.php");
exit;
?>

由于您的$_POST['names']和$_POST['cate']是数组，因此您无法将它们作为字符串进行检查，即。 if($_POST['names'] != '' && $_POST['cate'] == '') .此外，您正在嵌套循环，而您需要通过数组键链接它们。像——

foreach($_POST['names'] as $key => $val){
    if($_POST['names'][$key] != '' && $_POST['cate'][$key] == '') {
        $catego = mysqli_real_escape_string($con,$_POST['names'][$key]);
        $amo = mysqli_real_escape_string($con,$_POST['money'][$key]); 
        mysqli_query($con,"INSERT INTO cash (category, amount) VALUES ('".$catego."','".$amo."')");
    }
    else {
        $catego = mysqli_real_escape_string($con,$_POST['cate'][$key]);
        $amo = mysqli_real_escape_string($con,$_POST['money'][$key]); 
        mysqli_query($con,"INSERT INTO cash (category, amount) VALUES ('".$catego."','".$amo."')");
    }
}
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