Scripts

Datum uit form naar MYSQL

Door Ton -31-03-2006 00:00 -4.415 views

Met de 3 onderstaande bestandjes kun je in een formulier (waar ook de huidige datum reeds in staat) een datum met drop down menu selecteren. Deze kun je dan op een eenvoudige wijze aan je date veld in MYSQL toevoegen. Alleen de query zul je zelf erbij moeten zetten Enige wat nog aangepast zou kunnen worden is het invoeren van bv 31 feb. Dit zou niet mogelijk moeten zijn.

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datum-uit-form-naar-mysql

<?php 
    function DateSelector($inName, $useDate=0) 
    { 
        /* create array so we can name months */ 
        $monthName = array(1=> "januari", "februari", "maart", 
            "april", "mei", "juni", "Juli", "augustus", 
            "september", "oktober", "november", "december"); 
      
        /* if date invalid or not supplied, use current time */ 
        if($useDate == 0) 
        { 
            $useDate = Time(); 
        } 

        /* make day selector */ 
        echo "<SELECT NAME=" . $inName . "Day>\n"; 
        for($currentDay=1; $currentDay <= 31; $currentDay++) 
        { 
            echo "<OPTION VALUE=\"$currentDay\""; 
            if(intval(date( "d", $useDate))==$currentDay) 
            { 
                echo " SELECTED"; 
            } 
            echo ">$currentDay\n"; 
        } 
        echo "</SELECT>"; 

        /* make month selector */ 
        echo "<SELECT NAME=" . $inName . "Month>\n"; 
        for($currentMonth = 1; $currentMonth <= 12; $currentMonth++) 
        { 
            echo "<OPTION VALUE=\""; 
            echo intval($currentMonth); 
            echo "\""; 
            if(intval(date( "m", $useDate))==$currentMonth) 
            { 
                echo " SELECTED"; 
            } 
            echo ">" . $monthName[$currentMonth] . "\n"; 
        } 
        echo "</SELECT>"; 

        /* make year selector */ 
        echo "<SELECT NAME=" . $inName . "Year>\n"; 
        $startYear = date( "Y", $useDate); 
        for($currentYear = $startYear - 5; $currentYear <= $startYear+5;$currentYear++) 
        { 
            echo "<OPTION VALUE=\"$currentYear\""; 
            if(date( "Y", $useDate)==$currentYear) 
            { 
                echo " SELECTED"; 
            } 
            echo ">$currentYear\n"; 
        } 
        echo "</SELECT>"; 
     
    } 
?> 


form.php

[code]<?php include("dateselect.php"); ?>
<html>
<head>
</head>
<body>
<form method="post" action="add.php">
<TABLE>
<TR>
<TD>TEST:
<TD>
<input type=TEXT Name=TEST width=20>
</TD>
</TR>
<TR>
   <TD>datum:</TD>
   <TD>
      <!-- You can use PHP functions to automatically get the value of date -->
<?php DateSelector('day','month','year'); ?>
   </TD>
</TR>
<TR>
   <TD></TD><br>
   <TD><INPUT TYPE="submit" name="submit" value="submit"></TD> 
</TR>
</TABLE>
</form>

<body>
<html>[/code]

add.php

[code]<?php 
//initilize PHP
{
   //convert all the posts to variables:
   $message = $_POST['message'];
   $TEST = $_POST['TEST'];
   $tijd = $_POST['tijd'];
   $Dag = $_POST['dayDay'];
   $Maand = $_POST['dayMonth'];
   $Jaar = $_POST['dayYear'];
   // Dit is date VELD in je tabel
   $Datum = $Jaar."-".$Maand."-".$Dag; 
}
?>
   
<table width=50%>
<tr><td>Jaar: </td><td> <?php echo $Jaar; ?> </td></tr>
<tr><td>Dag: </td><td> <?php echo $Dag; ?> </td></tr>
<tr><td>Maand: </td><td> <?php echo $Maand; ?> </td></tr>
<tr><td>Datum: </td><td> <?php echo $Datum; ?> </td></tr>
<tr><td>Test: </td><td> <?php echo $TEST; ?> </td></tr>
</table>
[/code]