Question

我正在显示MySQL数据库中的表。我想在每列中使用pulldown来过滤表。我想在不使用提交按钮的情况下动态完成。

正如您将从代码和图片中看到的那样，我几乎设法做到这一点，但$_POST中返回的数据令人困惑。

首先：$_POST[name]，$_POST[car]和$_POST[color]都返回第一个值，无论是否选择了该选择功能。为什么是这样？我已编码（如你所见）

第二：如果我选择了Bob＆＃39;，代码第一次运行，但如果我选择“Bob＆＃39;再次，$_POST表现得像空。啊？

我认为基本代码的错误是什么？我在这个网站和互联网上看起来很高和很低，试图找到关于$ _POST如何在这方面发挥作用的更详细的解释。

<?php
//connect to database
$mysqli = mysqli_connect('localhost', 'root', '', "masterlist");

//get the column names from the masterlist table
$get_list_sql = "SHOW COLUMNS FROM listTest";
$get_list_res = mysqli_query($mysqli, $get_list_sql) or die(mysqli_error($mysqli));

//display_block contains the HTML form data to display
$display_block = "<h1>QLD RNE Design Master List</h1>";
$display_block .="<form action='listTest.php' method='POST'>";
$display_block .="<table><tr>";

//.............................Display field names as header..................................................................

//loop through all the fields, ignore the id field as we don't display it
  while ($list = mysqli_fetch_array($get_list_res)) {
    if ($list['Field'] <> 'id') {

        // simplify the data by using a string to store the current field name
        $Field = $list['Field'];

        // build select for the field
        $display_block .= "<td align = 'center'>".$Field."</td>";
    }
}
$display_block .="</tr><tr>";

//.............................Display Sorting options..................................................................

//loop through all the fields, ignore the id field as we don't display it
$get_list_res = mysqli_query($mysqli, $get_list_sql) or die(mysqli_error($mysqli));

$get_list_sql = "SELECT * from listTest order by name";
  while ($list = mysqli_fetch_array($get_list_res)) 
  {
    if ($list['Field'] <> 'id') 
    {
        // get the unique list of data in the field to use in the select field for filtering
        $SQL = "SELECT DISTINCT ".$list['Field']." from listTest ORDER BY ".$list['Field']." ASC ";
        $get_list = mysqli_query($mysqli, $SQL) or die(mysqli_error($mysqli)) ;  

        // simplify the data by using a string to store the current field name
        $Field = $list['Field'];

        if(isset($_POST[$Field]) && $_POST[$Field] <> $Field)
        {
            echo($Field.' should have been pressed and has a value of '.$_POST[$Field].'<br>');
            $get_list_sql = "SELECT * from listTest where ".$Field." ='" .$_POST[$Field]. "' order by name";
            $name = $_POST[$Field];
        }else
        {
            $name = $Field;
        }
        echo ('name = '.$name.'<br>');

        // build select for the field
        $display_block .= "<td align = 'center'><select name ='".$name."' id='".$Field."' onchange='this.form.submit()'>";

        //set the first option to the field name
        $display_block .= "<option value=".$Field.">".$Field."</option>";

        //loop through the data list
        while ($select = mysqli_fetch_array($get_list)) 
        {
            if ($select[$Field]<>null)
            {
                $display_block .= "<option value=".$select[$Field].">".$select[$Field]."</option>";
            }
        }
        $display_block .="</select></td>";
    }
}

//.............................Display Data..................................................................

foreach ($_POST as $key => $value) 
{
        echo "<pre><tr>";
        echo "<td> ";
        echo $key;
        echo "</td>";
        echo "<td> ";
        echo $value;
        echo "</td>";
        echo "</tr></pre>";
}

echo($get_list_sql );

$get_list_res = mysqli_query($mysqli, $get_list_sql) or die(mysqli_error($mysqli));
  while ($list = mysqli_fetch_array($get_list_res)) 
{
    $display_block .= "<tr><td>".$list['name']."
        </td><td>".$list['color']."
        </td><td>".$list['car']."
        </td>";
}
//echo('finished');
$display_block .="</tr>";
$display_block .="</form></table>";

//free results
mysqli_free_result($get_list_res);

//close connection to MySQL
mysqli_close($mysqli);
//$_POST = array();


//.............................HTML.................................................................
?>
<!DOCTYPE html>
<html>
<link rel="stylesheet" href="w3.css">
<head>
<style>
h1 {
    border-bottom: 3px solid #cc9900;
    color: #996600;
    font-size: 20px;
}
table, th , td {
    border: 1px solid grey;
    border-collapse: collapse;
    padding: 5px;
    font-size: 15px;

}
table tr:nth-child(odd) {
    background-color: #f1f1f1;
}
table tr:nth-child(even) {
    background-color: #ffffff;
}
</style>
</head>

<body>

<?php echo $display_block; ?>
</body>
</html>

由于我只能粘贴12个链接，这里是运行代码的一个例子。拳头图像是开始，第二个是选择＆＃39; bob＆＃39;第三是选择＆＃39; bob＆＃39;第二次：

我希望有人可以提供帮助，请不要建议我用其他语言写这个。我希望它在PHP中。我使用Chrome作为网络浏览器

$ _POST没有按照我的预期进行

0 个答案: