php - 在多个表中搜索记录，并在mysql和php中以不同的方式输出输出

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

所有4个表都有一个列PID，我要从中进行用户查询。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，然后在TABLE2中搜索，依此类推。

由于所有表都具有不同的结构，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，则输出将有一个包含5列的表，或者如果从TABLE4中得到的输出将显示11列。

现在，我只能查询一张桌子。

<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Search.php的输出表。

<?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

最佳答案

$query = "select * from TABLE1 where PID like '$_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/>"; ...
       }

关于php - 在多个表中搜索记录，并在mysql和php中以不同的方式输出输出，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/32884535/