每次单击按钮后如何从mysql数据库中逐行获取价值？
[英]How to get value from mysql database row by row after each click on button?
问题描述
I want to get questions one by one by clicking on button from mysql database table. 
我想通过单击mysql数据库表中的按钮来一个个地回答问题。 Each question is one single row, want to get every next question by clicking on "next" button from every next row. 每个问题都是一行，希望通过单击每一行的“下一个”按钮来获得下一个问题。 I have made this code but this only shows first question in the table of first row in the database. 我已经编写了此代码，但这仅在数据库第一行的表中显示了第一个问题。 My html code is: 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>
php code getting value from database is: 从数据库中获取价值的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']
);
$number++;
}
2 个解决方案
解决方案1
0 已采纳 2014-08-27 10:09:38
<html>
<head>
<style>
.invisible{
display:none;
}
.visible{
display:visible;
}
</style>
<script src="js/jquery.min.js"></script>
<script>$(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>
Something I put together real quick. 我快速整理的东西。
解决方案2
0 2014-08-28 07:13:25
$("#nex").click(function() { var formData1 = $("#ques").val(); $（“＃nex”）。click（function（）{var formData1 = $（“＃ques”）。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 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); 回声json_encode（$ record）;
mysqli_close($con); mysqli_close（$ con）; ?> ？>
Facing some problem in retreiving json data 检索json数据时遇到一些问题
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