Question

每个申请人都有自己的代码，该代码发生在“ applicant_code”中，并且可能有1个或多个受抚养人。该表中的主键是'dep_id'，它会自动递增。我想更改一行中一个字段的值，但显然它只会更新最后一行。

第一行不更新，仅第二行。我这个问题已经困扰了近一个星期了。请帮助我:(我认为查询只执行一次。

<?php 

include 'php/connection.php';

if (isset($_POST['edit'])) {

    $txtDependentID = $_POST["txtDependentID"];
    $txtDependentName = $_POST["txtDependentName"];
    $txtDependentRelation = $_POST["txtDependentRelation"];
    $txtDependentBirthDate = $_POST["txtDependentBirthDate"];
    $numdep = count($txtDependentName);

    for($i=0;$i<$numdep;$i++) {
        $sql = "UPDATE dependent_info SET dep_name='".$txtDependentName[$i]."', dep_relation='".$txtDependentRelation[$i]."', dep_birthdate='".$txtDependentBirthDate[$i]."' WHERE applicant_code= 
    '".$_GET['id']."';";
    }

    if (mysqli_query($conn, $sql)) { 
        echo "<script type='text/javascript'>
            alert('Success!');
            window.location.replace('http://localhost/1/view.php');
           </script>";
    } else {
      echo "Error ".mysqli_error($conn);
    }

}

这是我的HTML文件，用于在文本字段内显示数据库中的值。

        <table class="table" id="dependent_dynamic_field">
          <tr>
             <th style="width: 500px">ID</th>
             <th style="width: 500px">Dependent Name</th>
             <th style="width: 400px">Relation</th>
             <th style="width: 305px">Birthdate</th>
             <th style="width: 60px"></th>
          </tr>


            <?php 

            $dep_name = '';
            $dep_relation = '';
            $dep_birthdate = '';


            $sql = "SELECT * FROM dependent_info;";
                $result = mysqli_query($conn, $sql);
                $resultCheck = mysqli_num_rows($result);

                if ($resultCheck > 0) {
                  while ($row = mysqli_fetch_assoc($result)) 
                    {

                      $dep_id = $row['dep_id'];
                      $dep_name = $row['dep_name'];
                      $dep_relation = $row['dep_relation'];
                      $dep_birthdate = $row['dep_birthdate'];

            ?>
                <tr>

                  <td style="width: 500px"><input type="text" name="txtDependentID"  value="<?=$dep_id?>" class="form-control name_list" /><span style="color: red" id="dependentname"></span></td>
                  <td style="width: 500px"><input type="text" name="txtDependentName[]" id="txtDependentName"  value="<?=$dep_name?>" class="form-control name_list" /><span style="color: red" id="dependentname"></span></td>
                  <td style="width: 400px"><input type="text" name="txtDependentRelation[]" id="txtDependentRelation" value="<?=$dep_relation?>" class="form-control name_list" /><span style="color: red" id="dependentrelation"></span></td>
                  <td style="width: 305px"><input type="date" name="txtDependentBirthDate[]"  id="txtDependentBirthDate" value="<?=$dep_birthdate?>" class="form-control name_list" /><span style="color: red" id="dependentbirthdate"></span></td>

                   <td style="width: 60px"><button type="button" name="add_more_dependent" disabled id="add_more_dependent" class="btn btn-warning"><span style="font-size: 14px;" class="glyphicon glyphicon-plus" ></span></button></td>
                </tr>

            <?php
                    }
                  }



             ?>


          </tr>
        </table>

这是我将它们插入数据库的方式。正常运行

  $number = count($_POST["txtDependentName"]);
            if($number > 0){
            for($i=0; $i<$number; $i++){

            $sql .= "INSERT INTO dependent_info (dep_name, dep_relation, dep_birthdate) VALUES(mysqli_real_escape_string($conn, $_POST["txtDependentName"][$i])."','".mysqli_real_escape_string($conn, $_POST["txtDependentRelation"][$i])."','".mysqli_real_escape_string($conn, $_POST["txtDependentBirthDate"][$i])."');";
            }

        }

数组的第一行未更新

0 个答案: