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php – 从服务器动态加载数据到“选择”数据表中的输入字段

发布时间:2020-12-13 22:50:56 所属栏目:PHP教程 来源:网络整理
导读:我有一个 PHP代码,由函数加载器调用,只在datatable中的onInitCreate事件上触发.我想要做的是,当用户点击添加按钮时,它必须在选择字段中加载教员名称. 我已经尝试过这段代码,但它没有返回任何内容,或者我应该说一个空值.我确信它应该只返回一行. 这是我的PHP
我有一个 PHP代码,由函数加载器调用,只在datatable中的onInitCreate事件上触发.我想要做的是,当用户点击添加按钮时,它必须在选择字段中加载教员名称.

我已经尝试过这段代码,但它没有返回任何内容,或者我应该说一个空值.我确信它应该只返回一行.
这是我的PHP代码,它获取教师的名字并将其返回.

getFacultyNames.php

<?php
error_reporting(-1);
require_once("config.php"); 


$sql="SELECT lastName,firstName,middleName FROM table_faculty";
$result = mysql_query($sql);
$stack=array();

if($result === FALSE) {
    die(mysql_error()); // TODO: better error handling
}

    while($row = mysql_fetch_array($result))
          {
            $name = $row[0].",".$row[1]." ".$row[2];
            array_push($stack,array("label" => $name,"value" => $name));
          }
echo json_encode($stack); //here it returns [{"label":"Last,First Middle","value":"Last,First Middle"}]
?>

jquery代码:

function loader(){
$.ajax({
  "url": 'php/getFacultyNames.php',"async": false,"dataType": 'json',"success": function (json) {

         console.log( json );
         //the getFacultyNames.php is now returning correct values,//but how would I be able to get the value of the json code properly?
         //it always throws an error ""parsererror" SyntaxError
         //is it proper to have a code `return json;` in this success function?
    },"error" : function( jqXHR,textStatus,errorThrown ){ console.log( jqXHR,errorThrown ); }

});
}

这是我的编辑器初始化代码:

var editor = new $.fn.dataTable.Editor( {
        "ajaxUrl": "php/table.facultyloading.php","domTable": "#facultyloading","events": {
            "onInitCreate":function (){
                editor.add( {
                    "label": "Assign to Faculty","name": "facultyName","type": "select","ipOpts":loader()      // Returns array of objects - .ajax() with async: false
                    });
                }
            },"fields": [
            {
                "label": "Subject Name","name": "name","ipOpts": [
                    {
                        "label": "sample","value": "sample"
                    }
                ]
            },{
                "label": "Day","name": "day","default": "Monday","type": "checkbox","ipOpts": [
                    {
                        "label": "Monday ","value": "Monday "
                    },{
                        "label": " Tuesday ","value": " Tuesday "
                    },{
                        "label": " Wednesday ","value": " Wednesday "
                    },{
                        "label": " Thursday ","value": " Thursday "
                    },{
                        "label": " Friday ","value": " Friday "
                    },{
                        "label": " Saturday","value": " Saturday"
                    }
                ],"separator": "|"
            },{
                "label": "Start Time","name": "startTime","type": "text"
            },{
                "label": "End Time","name": "endTime",{
                "label": "Room","name": "room","type": "text"
            }
        ]
    } );

我似乎无法弄清楚出了什么问题.我错过了什么?你能帮助我吗?
提前致谢!

解决方法

我已经将mysql函数转换为PDO_MySQL,最后它的工作原理是我的新getFacultyNames.php,我也修改了我的jquery代码.感谢你的帮助!

(编辑:李大同)

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