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Php 如何在codeigniter中插入从javascript到mysql的数组?_Php_Mysql_Arrays_Ajax - Fatal编程技术网

Php 如何在codeigniter中插入从javascript到mysql的数组?

Php 如何在codeigniter中插入从javascript到mysql的数组?,php,mysql,arrays,ajax,Php,Mysql,Arrays,Ajax,我有一个让我发疯的问题,那就是我需要记录视图表中输入的数据,我已经可以获得所有这些输入的数据,但我无法在数据库中注册它们 到目前为止,我可以获得如下数组中的输入数据: string(142) "[{"name":"txtrut","value":"123"},{"name":"txtnombre","value":"asdasd"

我有一个让我发疯的问题,那就是我需要记录视图表中输入的数据,我已经可以获得所有这些输入的数据,但我无法在数据库中注册它们

到目前为止,我可以获得如下数组中的输入数据:

string(142) "[{"name":"txtrut","value":"123"},{"name":"txtnombre","value":"asdasd"}, 
{"name":"txtrut","value":"324"},{"name":"txtnombre","value":"asdaasa"}]"
“我的表”所在的视图如下所示:

<div id="trabajadores" class="lightbox-basic zoom-anim-dialog mfp-hide">
        <div class="container">
           <div class="row">
              <button title="Close (Esc)" type="button" class="mfp-close x-button">×</button>
              <div class="col-lg-4" id="dynamic_field">
                 <h3>Agregar personal</h3>
                 <hr>
                 <table class="table table-bordered" id="tabla_personal">
                    <TR>
                       <TH>Rut</TH>
                       <TH>Nombre completo</TH>
                       <TH></TH>
                    </TR>
                    <tr>
                        <TD><input type="text" name="txtrut" placeholder="Ingrese" class="form- 
                          control" required/></TD>
                       <TD><input type="text" name="txtnombre" placeholder="Ingrese" class="form-control" required/></TD>
                       <TD><button type="button" name="add" id="agregarTrabajador" class="btn btn-success">+</button></TD>
                    </tr>
                 </table>
              </div>
              <!-- end of col -->
              <!-- href="<?php echo base_url()?>PlantillaOperaciones"-->
              <a class="btn-solid-reg mfp-close page-scroll reg-trabajo" class="nav- 
             link">Siguiente</a>
           </div>
           <!-- end of row -->
        </div>
        <!-- end of container -->
     </div>
但问题是,我希望遍历来自ajax的$data变量,并以某种方式遍历它以获取数据,并能够将其注册到数据库中

public function registrarPersonal($data){
    //Registro  de personal
    */
    $listadopersonal = json_encode($data['listadopersonal']);
    var_dump($listadopersonal);
    $id_trabajodiario = 1;

    foreach ($listadopersonal as $key => $value){
        $querypersonal = "INSERT INTO personal (Rut,NombreCompleto,ID_TrabajoDiario) ". 
        "VALUES ('".$value."','". $listadopersonal[$value]."','".$id_trabajodiario."')"; 
    }

    return $this->db->query($querypersonal);
}
此时,变量$listadopersonal的值为:

string(142) "[{"name":"txtrut","value":"123"},{"name":"txtnombre","value":"asdasd"}, 
{"name":"txtrut","value":"324"},{"name":"txtnombre","value":"asdaasa"}]"
没有json_encode函数,它是:

array(2) { [0]=> array(2) { ["name"]=> string(6) "txtrut" ["value"]=> string(2) "12" } [1]=> array(2) 
{ ["name"]=> string(9) "txtnombre" ["value"]=> string(6) "asdasd" } }
这就是问题所在,如何将数据输入到数据库中我的个人表中,同时考虑到它可以是多个insert,也可以是多个insert。我知道codeigniter中有一个insert_批,但我正在寻找如何实现它,我非常感谢您阅读了所有这些内容,希望您能及时提供帮助

问候你,祝你有愉快的一天

string(142) "[{"name":"txtrut","value":"123"},{"name":"txtnombre","value":"asdasd"}, 
{"name":"txtrut","value":"324"},{"name":"txtnombre","value":"asdaasa"}]"
array(2) { [0]=> array(2) { ["name"]=> string(6) "txtrut" ["value"]=> string(2) "12" } [1]=> array(2) 
{ ["name"]=> string(9) "txtnombre" ["value"]=> string(6) "asdasd" } }