Question

我更新php表单并更改值，但它无法更改它保存相同的先前值

<tr>
  <td width="22%">Course Fees</td>
  <td width="38%">
    <div id="total_fees">
      <input type="text" class="input_text" name="total_fees" id="toal_fees"  onblur="show_record(this.value,1)" value="<?php if($_POST['total_fees']) echo $_POST['total_fees']; else echo $row['total_fees'];?>" />
    </div>
  </td>
</tr>

请参阅源代码和输入文件的屏幕截图。我更改了输入字段中的值，但它在源代码中保持相同并保存源代码值db

Answer 1

输入值不会立即影响您的输入，但您在提交表单时会获得新值。

然后提交表单，然后打印请求的值。您将在请求参数中找到total_fees的新值。

Answer 2

HTML CODE

<form name="jqueryForm" method="post" id="jqueryForm"  enctype="multipart/form-data" onSubmit="return validme();">

    <table border="0" cellspacing="15" cellpadding="0" width="100%" >
    <tr>
                        <td width="22%">Course Fees</td>
                        <td width="38%">
                        <div id=total_fees>

                        <input type="text" class="input_text" name="total_fees" id="toal_fees"  onblur="show_record(this.value,1)" value="<?php if($_POST['total_fees']) echo $_POST['total_fees']; else echo $row['total_fees'];?>" />
                        </div>
                        </td>                
                        <td width="40%"></td>
                    </tr>
                    <tr>
                        <td width="22%">Discount in <input type="radio" name='discount_type' value="<?php if($_POST['discount_type']) echo $_POST['discount_type']; else echo percent; ?>" checked="checked" onChange="show_record()" />% or <input type="radio" name='discount_type' value="<?php if($_POST['discount_type']) echo $_POST['discount_type']; else echo cash; ?>" onChange="show_record()" />Cash</td>
                        <td width="38%"><input type="text" class="input_text" name="concession" id="concession" value="<?php if($_POST['concession']) echo $_POST['concession']; else if($row_record['discount'] !=0) echo $row_record['discount']; else echo 0; ?>" onBlur="show_record()"/>
                        </td>
                        <td width="40%"></td>
                    </tr>
    </table>
    <input type="submit" value="Save Record" name="save_changes" />
    </form>

用于保存数据库的PHP代码

if(isset($_POST['save_changes']))
{
$total_fees=($_POST['total_fees']) ? $_POST['total_fees'] : "0";
$data_record['total_fees']=$total_fees;
$courses_id=$db->query_insert("enroll", $data_record);  
}

更新在PHP无法正常工作

2 个答案: