Question

我一直在玩PHP和ajax，试图让它产生我的结果。但出于某种原因，它根本不起作用。我会写一些字，什么都不会出现。

代码：

tabLayout.setOnTabSelectedListener(new TabLayout.OnTabSelectedListener() {
        @Override
        public void onTabSelected(TabLayout.Tab tab) {


            CharSequence _header = tab.getText();
            Selected_Tab_Header = String.valueOf(_header);

            if (Selected_Tab_Header == "Tab1") {
                headerImage.setImageResource(R.drawable.someImage);

            }

        }

        @Override
        public void onTabReselected(TabLayout.Tab tab) {

        }

        @Override
        public void onTabUnselected(TabLayout.Tab tab) {
        }
    });

同样，我不确定如何解决这个问题。我的数据库名为<script type="text/javascript"> $( document ).ready( function() { $('.searchFunction').keyup( function( event ) { var search_term = $(this).attr('value'); }); </script>，字段名称为ing。

Answer 1

知道了：你没有在var search_term = $(this).val();

获取文本框值

OP尝试使用var search_term = $(this).attr('value');

读取文本框的值

您的选择查询中存在问题，我已对其进行了更新。

更新1：

<?php
    if( $_SERVER['REQUEST_METHOD']=='POST' )
    {
        $conn=mysql_connect("localhost","root","") or die ("could not connect");
        mysql_select_db("test") or die ("could not find db");

        if ( !empty( $_POST['search_term'] ) ) 
        {

            $search_term =  $_POST['search_term'] ;

           $query = mysql_query( "SELECT `ingName` FROM `ing` WHERE `ingName` LIKE '".$search_term.'%', $conn );


            if( $query )
            {
                while( $row = mysql_fetch_assoc( $query ) ) 
                {
                    echo '<li>'.$row['ingName'].'</li>';
                }
            }
        }
        mysql_close( $conn );
        exit();
    }
?>

<!DOCTYPE html>
<html lang="en">
  <head>
    <title>Dashboard</title>
  </head>
  <body>

    <div class="container">
        <input type="text" name='search_term' class="searchFunction">
        <input type="submit" value="Search">
        <div class="dropdown">
            <ul class="result"></ul>
        </div>
    </div>

    <script src="http://code.jquery.com/jquery-2.2.1.min.js"></script>
    <script src="js/bootstrap.min.js"></script>
    <script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.3/jquery.min.js"></script>
    <script type="text/javascript">
        $( document ).ready( function() {
            $('.searchFunction').keyup( function( event ) {
                var search_term = $(this).val();

                $.post( document.location.href, { search_term:search_term }, function( data ) {
                    $('.result').html( data );

                    $('.result li').click( function( event ) {
                        var result_value = $(this).text();

                        $('.searchFunction').attr('value', result_value );
                        $('.result').html('');
                    });
                });
            });
        });
    </script>

  </body>

Answer 2

在MySQL命令行中运行查询，并尝试获取结果。你有回应吗？

下一点：你使用了错误的连接。我的意思是，改变这个

,$row['ingName'],

到这个

. $row['ingName'] .

Answer 3

按ID或类选择文本框。这里我使用id，在按下键时获取文本框的值，并将输入的值传递给ajax代码。

<?php
    if( $_SERVER['REQUEST_METHOD']=='POST' )
    {
        $conn=mysql_connect("localhost","root","") or die ("could not connect");
        mysql_select_db("test") or die ("could not find db");

        if ( !empty( $_POST['search_term'] ) ) 
        {

            $search_term = mysql_real_escape_string( $_POST['search_term'] );

            $query = mysql_query( "SELECT `ingName` FROM `ing` WHERE `ingName` LIKE '$search_term%'", $conn );


            if( $query )
            {
                while( $row = mysql_fetch_assoc( $query ) ) 
                {
                    echo '<li>'.$row['ingName'].'</li>';
                }
            }
        }
        mysql_close( $conn );
        exit();
    }
?>

<!DOCTYPE html>
<html lang="en">
  <head>
    <title>Dashboard</title>
  </head>
  <body>

    <div class="container">
        <input type="text" name='search_term' id="search_term" class="searchFunction">
        <input type="submit" value="Search">
        <div class="dropdown">
            <ul class="result"></ul>
        </div>
    </div>

    <script src="http://code.jquery.com/jquery-2.2.1.min.js"></script>
    <script src="js/bootstrap.min.js"></script>
    <script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.3/jquery.min.js"></script>
    <script type="text/javascript">
        $( document ).ready( function() 
        {
            $('.searchFunction').keyup( function( event ) 
            {
                var search_term = $("#search_term").val();

                $.post( document.location.href, { search_term:search_term }, function( data ) 
                {
                    $('.result').html( data );

                    $('.result li').click( function( event ) 
                    {
                        var result_value = $(this).text();
                        $('.searchFunction').attr('value', result_value );
                        $('.result').html('');
                    });
                });
            });
        });
    </script>
  </body>
</html>

PHP搜索功能不显示结果

3 个答案: