数据库里有记录,但查询响应的记录时,却显示“数据库查询失败!<br/>可能数据库中没有记录”,请教这是什么问题?搜索代码:
<form id="form1" name="form1" method="get" action="<?php echo $_SERVER['PHP_SELF']; ?>">
搜索:
<label>
<select name="mode" id="S_mode">
<option value="S_all" selected="selected">全部</option>
<option value="S_name">供应商</option>
<option value="S_birthday">生日</option>
<option value="S_qq">QQ</option>
<option value="S_mobile">手机</option>
<option value="S_email">邮箱</option>
<option value="S_address">地址</option>
</select>
</label>
<label>
<input name="word" type="text" id="S_word" />
</label>
<label>
<input type="submit" name="Submit" value="搜索" />
</label>
</form>查询代码:
<?php
$word = $_GET['word']; //获得关键字
$mode = $_GET['mode']; //获取搜索方式
if(!empty($word)) {
echo '<title>可以查询</title>';
echo '您查询的关键词是:“',$word,'”<br/>';
$sql = "SELECT * FROM ".SqlTableName;
switch($mode)
{
case 'S_all': $sql = 'SELECT * FROM `'.SqlDataBase.'`.`'.SqlTableName."`
WHERE `Name` LIKE '%$word%' OR `Birthday` LIKE '%$word%'
OR `QQ` LIKE '%$word%' OR `Mobile` LIKE '%$word%'
OR `Email` LIKE '%$word%' OR `Address` LIKE '%$word%'";break; case 'S_name': $sql = 'SELECT * FROM `'.SqlDataBase.'`.`'.SqlTableName."`
WHERE `Name` LIKE '%$word%'"; break;
case 'S_birthday': $sql = 'SELECT * FROM `'.SqlDataBase.'`.`'.SqlTableName."`
WHERE `Birthday` LIKE '%$word%'"; break;
case 'S_qq': $sql = 'SELECT * FROM `'.SqlDataBase.'`.`'.SqlTableName."`
WHERE `QQ` LIKE '%$word%'"; break;
case 'S_mobile': $sql = 'SELECT * FROM `'.SqlDataBase.'`.`'.SqlTableName."`
WHERE `Mobile` LIKE '%$word%'"; break;
case 'S_email': $sql = 'SELECT * FROM `'.SqlDataBase.'`.`'.SqlTableName."`
WHERE `Email` LIKE '%$word%'"; break;
case 'S_address': $sql = 'SELECT * FROM `'.SqlDataBase.'`.`'.SqlTableName."`
WHERE `Address` LIKE '%$word%''; break;
default: $sql = "SELECT * FROM `".SqlTableName."`";
}
require('conn.php'); //调用conn.php文件,执行数据库操作
$result = mysqli_query($conn,$sql) or die('数据库查询失败!<br/>可能数据库中没有记录'); //SQL语句在这里执行
if($result)
{
echo $result;
}
}
else
{
echo '<title>',$word,'</title>';
}
?>
<form id="form1" name="form1" method="get" action="<?php echo $_SERVER['PHP_SELF']; ?>">
搜索:
<label>
<select name="mode" id="S_mode">
<option value="S_all" selected="selected">全部</option>
<option value="S_name">供应商</option>
<option value="S_birthday">生日</option>
<option value="S_qq">QQ</option>
<option value="S_mobile">手机</option>
<option value="S_email">邮箱</option>
<option value="S_address">地址</option>
</select>
</label>
<label>
<input name="word" type="text" id="S_word" />
</label>
<label>
<input type="submit" name="Submit" value="搜索" />
</label>
</form>查询代码:
<?php
$word = $_GET['word']; //获得关键字
$mode = $_GET['mode']; //获取搜索方式
if(!empty($word)) {
echo '<title>可以查询</title>';
echo '您查询的关键词是:“',$word,'”<br/>';
$sql = "SELECT * FROM ".SqlTableName;
switch($mode)
{
case 'S_all': $sql = 'SELECT * FROM `'.SqlDataBase.'`.`'.SqlTableName."`
WHERE `Name` LIKE '%$word%' OR `Birthday` LIKE '%$word%'
OR `QQ` LIKE '%$word%' OR `Mobile` LIKE '%$word%'
OR `Email` LIKE '%$word%' OR `Address` LIKE '%$word%'";break; case 'S_name': $sql = 'SELECT * FROM `'.SqlDataBase.'`.`'.SqlTableName."`
WHERE `Name` LIKE '%$word%'"; break;
case 'S_birthday': $sql = 'SELECT * FROM `'.SqlDataBase.'`.`'.SqlTableName."`
WHERE `Birthday` LIKE '%$word%'"; break;
case 'S_qq': $sql = 'SELECT * FROM `'.SqlDataBase.'`.`'.SqlTableName."`
WHERE `QQ` LIKE '%$word%'"; break;
case 'S_mobile': $sql = 'SELECT * FROM `'.SqlDataBase.'`.`'.SqlTableName."`
WHERE `Mobile` LIKE '%$word%'"; break;
case 'S_email': $sql = 'SELECT * FROM `'.SqlDataBase.'`.`'.SqlTableName."`
WHERE `Email` LIKE '%$word%'"; break;
case 'S_address': $sql = 'SELECT * FROM `'.SqlDataBase.'`.`'.SqlTableName."`
WHERE `Address` LIKE '%$word%''; break;
default: $sql = "SELECT * FROM `".SqlTableName."`";
}
require('conn.php'); //调用conn.php文件,执行数据库操作
$result = mysqli_query($conn,$sql) or die('数据库查询失败!<br/>可能数据库中没有记录'); //SQL语句在这里执行
if($result)
{
echo $result;
}
}
else
{
echo '<title>',$word,'</title>';
}
?>
现在数据库里比如有“中华人民共和国AAA”这样一条供应商的记录,我选择S_mode为供应商,关键字为“AA”,然后点查询,结果啥都没有,一片空白如果我把其中的一个case条件改为:
case 'S_address': $sql = 'SELECT * FROM `mycg`.`providers` WHERE `Address` LIKE '%$word%''; break;
红色部分是实际的数据库名称,然后S_mode选“地址”,关键字随意,就会出现:2条警告信息:
1、Division by zero in XXX
2、mysqli_query() [function.mysqli-query]: Empty query in XXX
改成
$sql = “SELECT * FROM `mycg`.`providers` WHERE `Address` LIKE '%$word%'”
if($result)
{
echo $result;
}会不会这样写输出语句有问题?
改成 die("$sql<br>" . mysql_error());这样在出错时就可知道为什么错了
switch 写的太乱,不便于检查正确性
Catchable fatal error: Object of class mysqli_result could not be converted to string in
如果查不到,就没有任何输出,一片空白!
修改die以后,输出sql语句如下:
SELECT * FROM `$dbname`.`$tablename` WHERE `company` LIKE '%1%' OR `Birthday` LIKE '%1%' OR `QQ` LIKE '%1%' OR `Mobile` LIKE '%1%' OR `Email` LIKE '%1%' OR `tel` LIKE '%1%'
要想获取查询结果:
$arr = $result->fetch_all(MYSQLI_ASSOC);
var_dump($arr);
汗
算了我给你写吧
你该看看php怎么用mysql查询
while ($row = mysql_fetch_array($result, MYSQL_NUM)) {
printf ($row);
}
在这儿之前你先输出你的SQL,然后到数据库执行看看有结果没。这样你就知道是哪儿错了