[英]How to get value from mysql database row by row after each click on button?
我想通过单击mysql数据库表中的按钮来一个个地回答问题。 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。 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。 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']
);
$number++;
}
<html>
<head>
<style>
.invisible{
display:none;
}
.visible{
display:visible;
}
</style>
<script src="js/jquery.min.js"></script>
<script>$(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>
我快速整理的东西。
$(“#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
<?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'=> $row['answer']
);
}
回声json_encode($ record);
mysqli_close($ con); ?>
检索json数据时遇到一些问题
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