错误:绑定变量的数量与令牌的数量不匹配

问题描述

我只想在db(mySQL)中没有对应关系的情况下进行插入，但是他让我不是该语句.这是代码段

I want to make an insert only if there's no correspondence in the db (mySQL) but he makes me not the statement. Here's the snippet

    if ($sql->rowCount() > 0) {
        echo 'Non inserisci';
    } else {
        echo 'Inserisci';
        $db->beginTransaction();
        echo 'Ciao3';
        $sql = $db->prepare("INSERT INTO contatti (nome,cognome) VALUES (?,?)") or die('Ciao2');
        echo 'Ciao4';
        $sql->execute(array($_POST['nome'],$_POST['cognome']));
        echo 'Ciao5';
        $db->rollBack();
    }

SELECT在哪里

    $db->beginTransaction();
    $sql = $db->prepare("SELECT * FROM contatti WHERE nome = ? AND cognome = ? WHERE nome = ? AND cognome = ?") or die ('Ciao1');
    $sql->execute(array($_POST['nome'],$_POST['cognome']));
    $db->rollBack();

你能解释一下哪里出问题了吗?

Can you explain me where's the fault?

推荐答案

错误出在算术中

让我们计数令牌:

 SELECT * FROM contatti WHERE nome = ? // one
                       AND cognome = ? // two
                        WHERE nome = ? // three
                       AND cognome = ? // four

现在让我们计算绑定变量的数量:

now let's count number of bound variables:

array($_POST['nome'], // one
      $_POST['cognome']) // two

4显然不等于2.这就是问题

4 is apparently not equal to two. that's the problem

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