Question

我需要执行一个mysql查询，它将搜索（在列中）nome和cognome（在Prenotazioni表中）。

问题是：查询必须搜索名称和姓氏或姓氏和名称（它们也可以是3或4个值），我只有1个搜索表单。如果我有两个不同姓氏的约翰（黑色和白色），并且我将搜索表单放入约翰黑色或白色约翰，查询将给我两个结果

我如何解决它？

的index.php

<table border="10">
  <tr>
    <td align="center"><b>CERCA</b></td>
  </tr>
  <tr>
    <td>
      <table>
        <form method="post" action="./cerca.php">
        <tr>
          <td>Cerca:</td>
            <td width="250" align="left"><input type="text" autocomplete="off" size="70" name="ricerca"/>
          </td>
        </tr>                   
        <tr>
          <td></td>
    <td align="left"><input style="width:100px;" type="submit" value="Cerca"/>
        </tr>
        </form>
      </table>
    </td>
  </tr>
</table>

cerca.php

 <!doctype html><html>
<head>
<meta charset="utf-8">
<title>Ricerca</title>
</head>
<body>
<?php

include('./config.php');

$name = isset($_POST['ricerca']) ? preg_replace('/\s+/', ' ', trim($_POST['ricerca'])) : '';

if ($name != '') {
    $q = 'SELECT * FROM Prenotazioni WHERE 1=2';
    $parts = explode(' ', $name);

    foreach ($parts as $part) {
        $q .= ' OR nome LIKE \'%' . $part . '\' OR cognome LIKE \'%' . $part . '\'';
    }

    $exec = mysql_query($q);
    if (mysql_num_rows($exec)) {
        echo '<tr>
                <td width="242">
                    <table border="0" cellpadding="3" cellspacing="1" bgcolor="#CCCCCC">
                        <tr></tr>
                        <tr>
                            <td align="center" bgcolor="#FFFFFF"><strong>Nome</strong></td>
                            <td align="center" bgcolor="#FFFFFF"><strong>Cognome</strong></td>
                        </tr>';

        while($row = mysql_fetch_array($exec)) {
            echo '<tr>
                    <td align="center" bgcolor="#FFFFFF">' . $row['nome'] . '</td>
                    <td align="center" bgcolor="#FFFFFF">' . $row['cognome'] . '</td>                     <p><br>
                   <br>';
        }
    }
    else {
        echo 'Errore: nessuna corrispondenza trovata.';
    }

}
else {
    echo 'Errore: Inserisci un nome.';
}
?>

Answer 1

我在3天前遇到了同样的挑战，这解决了它：将cerca.php的第6-11行修改为此

$parts = explode(" ",$name);

$count = count($parts);

$q = "SELECT * FROM Prenotazioni WHERE";                     

for($i = 0;$i < $count;$i++)
{
    if($i != $count-1)
        $q = $q.
        " (nome LIKE '%".$parts[$i]."%' OR
          cognome LIKE '%".$parts[$i]."%') AND ";
    else
        $q = $q.
        " (nome LIKE '%".$parts[$i]."%' OR
          cognome LIKE '%".$parts[$i]."%')";
}
$exec = mysql_query($q);

鉴于＆＃34; john black ＆＃34;，＆＃34; black john ＆＃34;和＆＃34; bla joh ＆＃34;返回相同的结果和＆＃34;白约翰＆＃34;与＆＃34;黑约翰＆＃34;相同。我希望它可以帮到你

Answer 2

使用上面的mu代码创建mysql查询字符串并运行此查询并根据您的行生成

$q = "SELECT * FROM Prenotazioni WHERE";                     

for($i = 0;$i < $count;$i++)
{
    if($i) {
        $q .= ' and ';
    }
    $q = $q . " (concat(nome, ' ', cognome) LIKE '%'.$parts[$i].'%')";
}

Answer 3

修改好的代码：

$name = isset($_POST['ricerca']) ? preg_replace('/\s+/', ' ', trim($_POST['ricerca'])) : ''; if ($name != '') { $parts = explode(" ",$name); $count = count($parts); $q = "SELECT * FROM Prenotazioni WHERE"; for($i = 0;$i < $count;$i++) { if($i != $count-1) $q = $q. " (nome LIKE '%".$parts[$i]."%' OR cognome LIKE '%".$parts[$i]."%') AND "; else $q = $q. " (nome LIKE '%".$parts[$i]."%' OR cognome LIKE '%".$parts[$i]."%')"; } $exec = mysql_query($q); if (mysql_num_rows($exec)) { echo '<tr> <table border="0" cellpadding="3" cellspacing="1" bgcolor="#CCCCCC"> <tr></tr> <tr> <td align="center" bgcolor="#FFFFFF"><strong>Nome</strong></td> <td align="center" bgcolor="#FFFFFF"><strong>Cognome</strong></td> </tr>'; while($row = mysql_fetch_array($exec)) { echo '<tr> <td align="center" bgcolor="#FFFFFF">' . $row['nome'] . '</td> <td align="center" bgcolor="#FFFFFF">' . $row['cognome'] . '</td>'; } } else { echo 'Errore: nessuna corrispondenza trovata.'; } } else { echo 'Errore: Inserisci un nome.'; }

用php进行Mysql搜索查询

3 个答案: