Question

我想通过点击mysql数据库表中的按钮逐个问题。每个问题都是一行，想通过点击＆＃34; next＆＃34;来获得每个下一个问题。每一行的按钮。我已经制作了这段代码但这只显示了数据库第一行表格中的第一个问题。我的HTML代码是：

<span ng-repeat="record in records" id="next">  
 <p id="hello">{{record.ques_no}}.
{{record.question}}</p>
<p><input type="text" ng-model="ans" id="ans" value=""></p>
<p align="center"><a href="#next" id="nex" class="ui-btn ui-corner-all ui-btn-inline" onclick="">Next</a></p>
 </span>

从数据库获取值的PHP代码是：

$result=mysqli_query($con,"SELECT * FROM quest limit 1");
$record=array();
$number = 0;
while($row =mysqli_fetch_array($result))
{
 $record[] = array(
'ques_no'=> $row['ques_no'],
'question'=> $row['question'],
'answer'=> $row['answer']
);
$number++;
}

Answer 1

<html>
    <head>
    <style>
    .invisible{
    display:none;
    }

    .visible{
    display:visible;
    }
    </style>
    <script src="js/jquery.min.js"></script>
    <script>$(function() 
    {
        $( "#button" ).click(function()
        {
            $( "div.container div.invisible" ).first().addClass( "visible" ).removeClass("invisible");
        });
    });
    </script>
    </head>
    <div class="container">
        <div class="invisible">1</div>
        <div class="invisible">2</div>
        <div class="invisible">3</div>
        <div class="invisible">4</div>
        <div class="invisible">5</div>
        <div class="invisible">6</div>
        <div class="invisible">7</div>
        <div class="invisible">8</div>
        <div class="invisible">9</div>
        <div class="invisible">10</div>
    </div>
    <input type="button" id="button"/>
</html>

我快速拼凑的东西。

Answer 2

$（＆＃34;＃nex＆＃34;）。click（function（）{ var formData1 = $（＆＃34;＃ques＆＃34;）。val（）;

         $.ajax({
         type:'POST',
        data:{'tota':formData1},
        url:'list1.php',
        success:function(data){
            alert(data);
             var json = $.parseJSON(data);
             // $("#hello").html(data);
             $("#hello").html(json[0]);
        }
        });
        });

list1.php

<?php
$total=$_POST['tota'];
$input=1;
$con=mysqli_connect("localhost","root","root","school");
$result=mysqli_query($con,"SELECT * FROM quest LIMIT $total OFFSET $input");
$record=array();
echo $input;
while($row =mysqli_fetch_array($result))
{
   $record[] = array(
    'ques_no'=> $row['ques_no'],
    'question'=> $row['question'],
    'answer'=> $row['answer']

   );

}

echo json_encode（$ record）;

mysqli_close（$ CON）; ？＆GT;

面对retreiving json数据中的一些问题

如何在每次单击按钮后逐行从mysql数据库中获取值？

2 个答案: