从sql数据库检索数据并在表中显示 - 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

我已经创建了一个sql数据库（用phpmyadmin填充测量，我想从两个日期之间调用数据（用户通过输入HTML形式的FROM和TO日期选择DATE）和显示

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，在我的HTML表单下，有一些复选框，通过检查它们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.

每个复选框代表我的数据库的一列;因此与日期和小时列一起，显示任何被选中的内容（如果没有选中则显示所有内容）。

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，当我的复选框没有被选中时，显示所有内容，并且还可以管理其中一个复选框

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：我调用的数据显示两次。

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

问题：四个复选框。

我需要为每个可能的组合写一个sql查询，还是有更简单的方法？

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

尝试创建如下的复选框：

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'> 

尝试将您的PHP代码改为：

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，但如果您希望一个通用解决方案，您可以尝试使用此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 array。

and use the result to construct the $column_names array.

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