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我是新来的PHP,我想弄清楚如何比较两个mysql查询结果并回显出不同之处。基本上我有一个比较状态标志特征的数据库。用户将选择2个状态,我想要的是让php回显3组结果(仅状态1,共享特性和状态2)。我下面的粗略代码只反映了状态1和状态2的特征。有没有一种方法可以比较查询结果,以便我得到状态之间的差异以及共享特征。使用PHP来比较mysql结果,然后回显差异?
<?php $query1 = $_GET['query1'];
$query2 = $_GET['query2'];
$min_length = 2;?>
<!-- state 1 characteristics only -->
<div class="fluid s1charc">
<?php echo "<strong><p>Search results for $query1</strong></p>";?>
<?php if(strlen($query1) >= $min_length){$query1 = htmlspecialchars($query1);
$query1 = mysql_real_escape_string($query1);
$raw_results1 = mysql_query("SELECT * FROM charc_s WHERE `charc_f` ='$query1' ORDER BY FLAG_S") or die(mysql_error());
if(mysql_num_rows($raw_results1) > 0) {
while($results1 = mysql_fetch_array($raw_results1)){
echo "<table id='iseqchart' class='sortable'>
<tr valign='middle'>
<td ><a href='flag.php?FLAG_ID=".$results1['FLAG_ID']."'><img src='images/".$results1['IMAGE']."'></a></td>
<td> </td>
<td><strong><font size='3'><a href='flag.php?FLAG_ID=".$results1['FLAG_ID']."' style='color:#100783'>".$results1['FLAG_S']." (".$results1['STATE'].")</strong></a>
</td>
</tr>
</table>";
}
}
else{ // if there is no matching rows do following
echo "No results were found";
}
}
else{ // if query length is less than minimum
echo " ".$min_length;
}
?>
</div>
<!-- shared characteristics -->
<div class="fluid sharecharc">Shared</div>
<!-- state 2 characteristics only -->
<div class="fluid s2charc">
<?php echo "<strong><p>Search results for $query2</strong></p>"; ?>
<?php if(strlen($query2) >= $min_length){$query2 = htmlspecialchars($query2);
$query2 = mysql_real_escape_string($query2);
$raw_results = mysql_query("SELECT * FROM charc_s WHERE `charc_f` ='$query2' ORDER BY FLAG_S") or die(mysql_error());
if(mysql_num_rows($raw_results) > 0) {
while($results = mysql_fetch_array($raw_results)){
echo "<table id='iseqchart' class='sortable'>
<tr valign='middle'>
<td ><a href='flag.php?FLAG_ID=".$results['FLAG_ID']."'><img src='images/".$results['IMAGE']."'></a></td>
<td> </td>
<td><strong><font size='3'><a href='flag.php?FLAG_ID=".$results['FLAG_ID']."' style='color:#100783'>".$results['FLAG_S']." (".$results['STATE'].")</strong></a>
</td>
</tr>
</table>";
}
}
else{ // if there is no matching rows do following
echo "No results were found";
}
}
else{ // if query length is less than minimum
echo " ".$min_length;
}
?>
</div>
不知道这是你需要的,但它可能有帮助... http://php.net/manual/de/function.array-diff.php 这个函数会返回你不存在的每个值在两个阵列中。 –
使用@TanuelMategi建议的内容 - 您应该将结果保存在数组中,然后比较两者。 –