php - 在多个表中搜索记录并在 mysql 和 php 中以不同的方式打印输出

Question

我的数据库中有四个表。 TABLE1 有 5 列，TABLE2 有 7 列，TABLE3 有 9 列，TABLE4 有 11 列。

所有 4 个表都有一个列 PID，我想从中进行用户查询。即

SELECT pid FROM ....

我面临的问题是，如何从所有 4 个表中进行选择，如果在 TABLE1 中找不到我的查询，则在 TABLE2 中搜索，依此类推.

由于所有表都有不同的结构，因此结果必须以不同的方式传递给输出，例如，如果在 TABLE1 中找到查询，则输出将具有一个包含 5 列的表，或者如果来自 TABLE4 输出将显示 11 列。

目前，我只能查询 1 个表。

<h2>title</h2>
            <div class="panel panel-primary">
            <div class="panel-heading">SELECT NAME FROM LIST</div>
            <div class="panel-body">
            <form name="dropdown" action="http://search.php" method="post">
            <select class="form-control" data-style="btn-primary" name="p_ID">
<?php
//provide your hostname, username and dbname
$host="localhost"; 
$username="root";  
$password="";
$db_name="mydb"; 
$con=mysql_connect("$host", "$username", "$password")or die("cannot connect");
mysql_select_db("$db_name");

$sql = "select PID from TABLE1";
$result = mysql_query($sql);
while($row=mysql_fetch_array($result))
{
echo "<option value=$row[PID]>$row[PID]</option>";
}
?>

    </select><br><br>
            <button class="btn btn-primary center-block btn-lg" type="submit" >Search</button>
                </form> 

            </div>
        </div>
    </div>

这是来自输出表的 Search.php。

<?php   

//provide your hostname, username and dbname
$DB_HOST = 'localhost';
$DB_USER = 'root';
$DB_PASS = '';
$DB_NAME = 'mydb';
$mysqli = new mysqli($DB_HOST, $DB_USER, $DB_PASS, $DB_NAME);
if(!$mysqli) 
{
    echo 'Could not connect to the database.';
} 
else 
{
if(isset($_POST[PID]) )
{
//echo "Input by list<br>";

$query = "select * from TABLE1 where PID like '$_POST[PID]%' ";
$result = $mysqli->query($query) or die($mysqli->error.__LINE__);
    //$arr = array();
    if($result->num_rows <= 1) 
    {
    echo "<table class=\"table\" ><tr><th>Protein ID:</th>";
        while($row = $result->fetch_assoc()) 
        {
        echo "<td>$row[PID]</td><tr><th>Protein Domain:</th><td>$row[p_name]</td></tr>
        <tr><th>Interacting Proteins:</th><td>";
        $seqsplit1 = wordwrap($row[p_symbol_orf], 60, "-\n", true);
        echo "$seqsplit1</td>";

        echo "</tr><tr><th>Protein Sequence:</th><td><div>";
        $seqsplit = wordwrap($row[p_sequence], 60, "\n", true);
        echo ">$row[p_ts]<br>$seqsplit";
        echo "</div></td></tr><tr><th>Protein Length:</th><td>$row[p_length]</td></tr><tr><td colspan='2'>";

        }
        ?>          

<?php       

    echo "</td></tr></tbody></table>";
    }
    else
    {
        echo "No query found, please search next!";
    }
} 

这些是下表的结构

TABLE1
PID PNAME PSYMBOL PSEQ PLEN

TABLE2
PID PNAME1 PSYMBOL1 PNAME2 PSYMBOL2  PSEQ PLEN

TABLE3
PID PNAME1 PSYMBOL1 PNAME2 PSYMBOL2 PNAME3 PSYMBOL3 PSEQ PLEN

TABLE4
PID PNAME1 PSYMBOL1 PNAME2 PSYMBOL2 PNAME3 PSYMBOL3 PNAME4 PSYMBOL4 PSEQ PLEN

Answer 1

$query = "select * from TABLE1 where PID like '$_POST[PID]%'  ";
$result = $mysqli->query($query) or die($mysqli->error.__LINE__);
if($result->num_rows == 0){
    $query = "select * from TABLE2 where PID like '$_POST[PID]%'  ";
    $result = $mysqli->query($query) or die($mysqli->error.__LINE__);
}else if($result->num_rows == 0){
    $query = "select * from TABLE3 where PID like '$_POST[PID]%'  ";
    $result = $mysqli->query($query) or die($mysqli->error.__LINE__);
}else if($result->num_rows == 0){
    $query = "select * from TABLE4 where PID like '$_POST[PID]%'  ";
    $result = $mysqli->query($query) or die($mysqli->error.__LINE__);
}else if($result->num_rows == 0){
    $query = "select * from TABLE5 where PID like '$_POST[PID]%'  ";
    $result = $mysqli->query($query) or die($mysqli->error.__LINE__);
}

要处理表列计数冲突，应使用 foreach 语句

if($result->num_rows <= 1)    
   while($row = $result->fetch_assoc())
       foreach ($row as $r1=>$r){
           echo $r1.": ".$r."<br/>"; ...
       }