如何使用mysql_query（） [英] how to use mysql_query()

问题描述

下面是用于从表中搜索用户输入值并打印它的php代码。



header.php

  //   html代码 
 <  表格 >  
 < span class =code-keyword><   select     name   = 选择 >  
 <  选项    value   =  empty > 无<   / option  >  
 <  选项    value   =  小说 > 小说<   / option  >  
  <  选项    value   = 期刊 > 期刊< span class =code-keyword><   / option  >  
 <  选项    value   =  acadamic  >  acadamic <   / option  >  
 <   /选择 >  
 
 
 <  < span class =code-leadattribute> form     action   =     method   =  POST >  
 
 <  输入    type   =  text    value   =  在此处输入关键字...   名称  =  q    size   =  10    id   =  searchfield    title   =   searchfield    o nFocus   =  clearText（this）    onBlur   =  clearText（this）     /  >  
 <  输入    type   =  submit   名称  = 搜索    value   =     alt   = 搜索    id   =  searchbutton    title   = 搜索    /  >  
 
 
 <   / form  >  <   / form  < span class =code-keyword>>  
 // php code 
 <？ php 
 
  if （isset（$ _ GET ['  q ']）&&a熔点; $ _GET ['  q']！=  ）
 {
 if（$ _ GET ['  select'] ==   fiction）{
 
 
 $ result = $ _ GET ['  q'];  //  存储用户输入的值 
  //   echo $ a;  
 $ sqlcommand =   SELECT title，author FROM .USRBOOKS。  WHERE fiction ='$ result' ; 
 
}否则if（$ _ GET ['  select'] == < span class =code-string>  journals）{
 $ result = $ _ GET ['  q']; 
 $ sqlcommand =   SELECT title，author FROM .USRBOOKS。  WHERE fiction ='$ result'; 
 
} 
 
  else  if（$ _ GET [''  select'] ==   acadamic）{
 $ result = $ _ GET ['  q']; 
 $ sqlcommand =   SELECT title，author FROM .USRBOOKS。  WHERE fiction ='$ result'; 
 
} 
 
 
  $ query  = mysql_query（$ sqlcommand）或 die（$ sqlcommand）; 
  echo    hello world; 
  $ count  = mysql_num_rows（$ query）; 
 
  if （$ count> 0）
 {
  global   $ searchoutput ; 
  $ searchoutput  =   $ count results ; 
  echo  ' < META http-equiv =refreshcontent =0; URL =' .INDEX_URL。'  result.php> ';  //  尝试在新的php文件中打印结果 
} 
} 
 
  else  {
 $ searchoutput =   0 result; 
  echo  ' < META http-equiv =refreshcontent =0; URL =' .INDEX_URL。'  resultn.php> ; 
 
} 
 
 ？>  

当我试图搜索该值时，它正在打印选择查询。

如果我输入值xyz（假设它存在于虚构列的数据库表中）

而不是将资源存储到$ query（通过使用$ mysql_query）它打印低于值



SELECT title，author FROM books WHERE fiction = 'xyz'



请任何人解决我的问题

解决方案

_GET [' q']）&&

_GET [' q']！= ）
{
if（

_GET [' select'] == 小说）{

below is the php code for searching user entered value from table and printing it.

header.php

 //html code
<form>
 <select name="select">
<option value="empty">none</option>
<option value="fiction">fiction</option>
<option value="journals">journals</option>
<option value="acadamic">acadamic</option>
</select>


            <form action="" method="POST">
 
           <input type="text" value="Enter a keyword here..." name="q" size="10" id="searchfield" title="searchfield" onFocus="clearText(this)" onBlur="clearText(this)" />
                    <input type="submit" name="Search" value="" alt="Search" id="searchbutton" title="Search" />


            </form></form>
// php code
<?php

if (isset($_GET['q']) && $_GET['q'] != "")
{
if($_GET['select'] == "fiction"){

  
   $result=$_GET['q'];// storing user entered value
   //echo $a;
   $sqlcommand= "SELECT title,author FROM ".USRBOOKS." WHERE fiction = '$result'";

}else if($_GET['select'] == "journals"){
$result=$_GET['q'];
	 $sqlcommand= "SELECT title,author FROM ".USRBOOKS." WHERE fiction = '$result'";

}

else if($_GET['select'] == "acadamic"){
$result=$_GET['q'];
 $sqlcommand= "SELECT title,author FROM ".USRBOOKS." WHERE fiction = '$result'";

}
 

$query = mysql_query($sqlcommand) or die($sqlcommand);
echo "hello world";
$count = mysql_num_rows($query);

if ($count>0)
{
global $searchoutput;
$searchoutput = " $count results ";
echo '<META http-equiv="refresh" content="0;URL='.INDEX_URL.'result.php">';// trying to print result in new php file
}
}

else {
 $searchoutput="0 result";
echo '<META http-equiv="refresh" content="0;URL='.INDEX_URL.'resultn.php">';

}

?>

when ever I am trying to search the value, its is printing select query.
if I entered value xyz (assume it is present in data base table in fiction column)
instead of storing resource into $query (by using $mysql_query) it is printing below value

SELECT title,author FROM books WHERE fiction = 'xyz'

please anyone solve my problem

解决方案

_GET['q']) &&

_GET['q'] != "") { if(

_GET['select'] == "fiction"){

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