我需要一个表格中的下拉菜单,才能将一些数据从PHP更新到mysql

时间:2019-01-09 16:06:44

标签: php mysql

我有一个带有一些左外部联接表的MySql查询和一个将编辑和更新MySql表的php表 我为属于我正在编辑的表的字段输入了一些文本。我也有一些现在是只读的文本输入,因为它们是MySql表中的外键,但是我需要在php表中将它们看成文本才能知道数据。我想要的是能够同时编辑php表中的那些字段,并且需要下拉列表而不是文本输入:     1-他们需要默认当前数据,我希望能够从下拉列表中进行选择以将当前数据更改为其他数据。如果需要的话     *默认值应来自我要更新的表,但选择选项应来自外部表     2-而且我需要将该选择作为外键(整数)传递,以便能够更新MySql表

 <body>

    <!-- this is my table header-->        
        <table>
    <thead>
    <tr>

                <th>Transaction Category</th>
                <th>Transaction Group</th>

                <th>Transaction Name</th>
                <th>Amount</th>
                <th>Date from</th>
                <th>Date to</th>
                <th>Notes</th>



        </tr></thead>


        <?php



 //this is my form/Table      
if (mysqli_num_rows($result2) > 0 ){


        while($row = mysqli_fetch_assoc($result2)){
 echo '<tr><form method="post">';



        echo  '<td><input type="text" name="transCat['.$row["budgetEstimatedId"].']" value="'.$row["transCat"].'" readonly="readonly"></td>';


       echo  '<td><input type="text" name="transGroup['.$row["budgetEstimatedId"].']" value="'.$row["transGroup"].'" readonly="readonly"></td>';
        echo '<td><select name="transGroup['.$row["budgetEstimatedId"].']" >'; 
                while($row1 = mysqli_fetch_assoc($result3)){ 
               echo '<option value='.$row1["transGroupsId"].'" if('.$row1["transGroup"].'=='.$row1["transGroupsId"].' selected="selected")>'.$row1['transGroup'].'</option>';
                   }
              echo '</select></td>';
        echo  '<td><input type="text" name="transNames['.$row["budgetEstimatedId"].']" value="'.$row["transNames"].'" readonly="readonly"></td>';

         echo  '<td><input type="decimal" name="amount['.$row["budgetEstimatedId"].']" value="'.$row["amount"].'"> </td>';

            echo  '<td><input type="date" name="dateFrom['.$row["budgetEstimatedId"].']" value="'.$row["dateFrom"].'"></td>';

            echo  '<td><input type="date" name="dateTo['.$row["budgetEstimatedId"].']" value="'.$row["dateTo"].'"> </td>';

            echo  '<td><textarea name="notes['.$row["budgetEstimatedId"].']"cols=34 rows=2 wrap=soft>'.$row["notes"].'</textarea> </td>';



         echo   '<td><input type="hidden" name="budgetEstimatedId[]" value="'.$row["budgetEstimatedId"].'"></td>';

        }   

    echo '<input type="submit" name="submit" value="Multiple CHANGE">';
        echo '<td><input type="submit" name="submit" value="Multiple CHANGE">';
    echo '</form></tr>';

        }

    ?>


        </table>    

 </body>
    <?php
       //this is where the insert start
if(isset($_POST["submit"]))
    {

foreach($_POST["budgetEstimatedId"] AS $budgetEstimatedIdID){
    $amountID = $_POST["amount"][$budgetEstimatedIdID];
$dateFromID = $_POST["dateFrom"][$budgetEstimatedIdID];
$dateToID = $_POST["dateTo"][$budgetEstimatedIdID];
$notesID = $_POST["notes"][$budgetEstimatedIdID];





$update = "UPDATE budgetestimated SET amount='$amountID', dateFrom='$dateFromID', dateTo='$dateToID', notes='$notesID' WHERE budgetEstimatedId='$budgetEstimatedIdID' LIMIT 1";
    mysqli_query($con, $update) or die (mysqli_error());
    }

    }
?>   

0 个答案:

没有答案