从 sql 数据库中检索数据并显示在表中 - 根据选中的复选框显示某些数据 [英] Retrieve data from sql database and display in tables - Display certain data according to checkboxes checked

问题描述

我创建了一个 sql 数据库(使用 phpmyadmin)，其中填充了测量值，我想从中调用两个日期之间的数据(用户通过在 HTML 表单中输入FROM"和TO"日期来选择日期)并显示他们在一张桌子上.

I have created an sql database(with phpmyadmin) filled with measurements from which I want to call data between two dates( the user selects the DATE by entering in the HTML forms the "FROM" and "TO" date) and display them in a table.

此外，我在我的 html 表单下放置了一些复选框，通过选中它们，您可以限制显示的数据量.

Additionally I have put, under my html forms, some checkboxes and by checking them you can restrict the amount of data displayed.

每个复选框代表我数据库的一列；因此，连同日期和小时列，会显示所有选中的内容(如果未选中任何内容，则显示所有内容).

Each checkbox represent a column of my database; so along with the date and hour column, anything that is checked is displayed(if none is checked then everything is displayed).

到目前为止，我设法编写了一个连接到数据库的 php 脚本，在没有选中任何复选框时显示所有内容，并设法将我的一个复选框排序.

So far I managed to write a php script that connects to the database, display everything when none of my checkboxes is checked and also managed to put in order one of my checkboxes.

问题:我调用的数据显示了两次.

Problem: The data that I call for are been displayed twice.

问题:我想要四个复选框.

Question: I want to have four checkboxes.

我是否需要为每种可能的组合编写一个 sql 查询，或者有更简单的方法吗?

Do I need to write an sql query for every possible combination or there is an easier way?

<?php
# FileName="Connection_php_mysql.htm"
# Type="MYSQL"
# HTTP="true"
$hostname_Database_Test = "localhost";
$database_Database_Test = "database_test";
$table_name = "solar_irradiance";
$username_Database_Test = "root";
$password_Database_Test = "";
$Database_Test = mysql_pconnect($hostname_Database_Test, $username_Database_Test,  $password_Database_Test) or trigger_error(mysql_error(),E_USER_ERROR); 


//HTML forms -> variables
$fromdate = $_POST['fyear'];
$todate = $_POST['toyear'];

//DNI CHECKBOX + ALL
$dna="SELECT DATE, Local_Time_Decimal, DNI FROM $database_Database_Test.$table_name   where DATE>="$fromdate" AND DATE<="$todate"";
$tmp ="SELECT * FROM $database_Database_Test.$table_name where DATE>="$fromdate" AND DATE<="$todate""; 

$entry=$_POST['dni'];
if (empty($entry))
{
$result = mysql_query($tmp);
echo 
"<table border='1' style='width:300px'>
<tr>
<th>DATE</th>
<th>Local_Time_Decimal</th>
<th>Solar_time_decimal</th>
<th>GHI</th>
<th>DiffuseHI</th>
<th>zenith_angle</th>
<th>DNI</th>
";

while( $row = mysql_fetch_assoc($result))
{
echo "<tr>";  
echo "<td>" . $row['DATE'] . "</td>";   
echo "<td>" . $row['Local_Time_Decimal'] . "</td>";  
echo "<td>" . $row['Solar_Time_Decimal'] . "</td>";  
echo "<td>" . $row['GHI'] . "</td>";  
echo "<td>" . $row['DiffuseHI'] . "</td>";  
echo "<td>" . $row['Zenith_Angle'] . "</td>";  
echo "<td>" . $row['DNI'] . "</td>";  
echo "</tr>";
}

echo '</table>';}

else
{
$result= mysql_query($dna);
echo
"<table border='1' style='width:300px'>
<tr>
<th>DATE</th>
<th>Local_Time_Decimal</th>
<th>DNI</th>
";

while($row = mysql_fetch_assoc($result))
{
echo "<tr>";  
echo "<td>" . $row['DATE'] . "</td>";  
echo "<td>" . $row['Local_Time_Decimal']."</td>";
echo "<td>" . $row['DNI'] . "</td>";  
echo "</tr>";
}
echo '</table>';
}
if($result){
        echo "Successful";
    }
    else{
    echo "Enter correct dates";
    }
?>
<?php
mysql_close();
?>

推荐答案

尝试创建您的复选框，如下所示:

Try to create your checkbox like below:

Solar_Time_Decimal<checkbox name='columns[]' value='1'>
GHI<checkbox name='columns[]' value='2'>
DiffuseHI<checkbox name='columns[]' value='3'>
Zenith_Angle<checkbox name='columns[]' value='4'>
DNI<checkbox name='columns[]' value='5'> 

并尝试将您的 PHP 代码挂起:

And try to hange your PHP code to this:

<?php
//HTML forms -> variables
$fromdate = isset($_POST['fyear']) ? $_POST['fyear'] : data("d/m/Y");
$todate = isset($_POST['toyear']) ? $_POST['toyear'] : data("d/m/Y");
$all = false;
$column_names = array('1' => 'Solar_Time_Decimal', '2'=>'GHI', '3'=>'DiffuseHI', '4'=>'Zenith_Angle','5'=>'DNI');
$column_entries = isset($_POST['columns']) ? $_POST['columns'] : array();
$sql_columns = array();
foreach($column_entries as $i) {
   if(array_key_exists($i, $column_names)) {
    $sql_columns[] = $column_names[$i];
   }
}
if (empty($sql_columns)) {
 $all = true;
 $sql_columns[] = "*";
} else {
 $sql_columns[] = "DATE,Local_Time_Decimal";
}

//DNI CHECKBOX + ALL
$tmp ="SELECT ".implode(",", $sql_columns)." FROM $database_Database_Test.$table_name where DATE>="$fromdate" AND DATE<="$todate""; 

$result = mysql_query($tmp);
echo "<table border='1' style='width:300px'>
<tr>
<th>DATE</th>
<th>Local_Time_Decimal</th>";
foreach($column_names as $k => $v) { 
  if($all || (is_array($column_entries) && in_array($k, $column_entries)))
     echo "<th>$v</th>";
}
echo "</tr>";
while( $row = mysql_fetch_assoc($result))
{
    echo "<tr>";  
    echo "<td>" . $row['DATE'] . "</td>";   
    echo "<td>" . $row['Local_Time_Decimal'] . "</td>";  
    foreach($column_names as $k => $v) { 
      if($all || (is_array($column_entries) && in_array($k, $column_entries))) {
         echo "<th>".$row[$v]."</th>";
       }
    }
    echo "</tr>";
}
echo '</table>';

if($result){
        echo "Successful";
    }
    else{
    echo "Enter correct dates";
    }
?>
<?php
mysql_close();?>

此解决方案考虑您的特定表列，但如果您希望通用解决方案，您也可以尝试使用此 SQL:

This solution consider your particular table columns but if your wish a generic solution you can try to use this SQL too:

$sql_names = "SELECT COLUMN_NAME FROM INFORMATION_SCHEMA.COLUMNS WHERE TABLE_SCHEMA = '$database_Database_Test' AND TABLE_NAME = '$table_name'";

并使用结果构造$column_names数组.

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