Question

我需要将来自后调用的插入更新为MYSQL DB中的相同数据。这是表单代码：

<form action="salva_debiti.php" method="post">


'; // Create connection $conn = new mysqli($servername, $username, $password, $dbname); // Check connection if ($conn->connect_error) { die("Connection failed: " . $conn->connect_error); } $sql = "SELECT * FROM rimanenti_mese WHERE Mese = $mese AND Anno = $anno ORDER BY cognome ASC"; $result = $conn->query($sql); if ($result->num_rows > 0) { echo '
<table id="ricambi" style="width: 100%;" border="1">
    <tr>
        <td id="titolotd">Nome</td>
        <td id="titolotd">Cognome</td>
        <td id="titolotd">Da incassare pezzi</td>
        <td id="titolotd">Da incassare benzina</td>
        <td id="titolotd">Residuo '.$mese_scritto.' '.$anno_pr_div.'</td>
    </tr>'; while($row = $result->fetch_assoc()) { echo '

    <tr>

        <td>
            <input type="" name="nome[]" value="'.$nome.'">
        </td>
        <td>
            <input type="" name="cognome[]" value="'.$cognome.'">
        </td>
        <td>
            <input type="" name="pezzi[]" value="'.$pezzi.'">
        </td>
        <td>
            <input type="" name="benzina[]" value="'.$benzina.'">
        </td>
        <td>
            <input type="" name="rimanenti[]" value="'.$rimanenti.'" </td>
    </tr>
    ' ; } echo '
    <input type="submit" value="Submit">'; echo '</form>';

这是我用来插入数据的代码：

// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn - > connect_error) {
die("Connection failed: ".$conn - > connect_error);
}

 foreach($_POST['nome'] as $nome) {

$nome = $_POST['nome'];
$cognome = $_POST['cognome'];
$rimanenti_ins = $_POST['rimanenti'];


//$sql = "UPDATE rimanenti_mese SET Rimanenti='$rimanenti_ins' WHERE Nome='$nome' AND Cognome = '$cognome'";

$sql = "INSERT INTO rimanenti_mese (Nome, Cognome, Mese, Anno, Benzina, Pezzi, Rimanenti)
VALUES('$nome', '$cognome', '09', '2015', 'xxxx', 'fdfdfd', '$rimanenti_ins')
";

}

 if ($conn - > query($sql) === TRUE) {
echo "New record created successfully";
} else {
echo "Error: ".$sql.
"<br>".$conn - > error;
}

 $conn - > close();

问题是它只插入最后一行...... 我哪里错了？

Answer 1

将$conn - > query($sql)移至foreach圈。

foreach($_POST['nome'] as $nome) {
    $nome = $_POST['nome'];
    $cognome = $_POST['cognome'];
    $rimanenti_ins = $_POST['rimanenti'];
    $sql = ".....";
    if ($conn - > query($sql) === TRUE) {

    } else {

    }
}

因为现在您只进行了一次查询，并且在foreach循环结束后$sql包含查询最后一项。

<强>更新：单词Array是数组值的字符串表示。执行$nome = $_POST['nome']时，$_POST['nome']是数组，因此将数组添加到字符串会生成单词Array。因此，您应该在每次迭代时获取数组的每个项目，例如：

$i = 0;
foreach ($_POST['nome'] as $nome) {
    $nome = $_POST['nome'][$i];
    $cognome = $_POST['cognome'][$i];
    $rimanenti_ins = $_POST['rimanenti'][$i];

    // sql
    // run query

    $i++;
}

Answer 2

您在foreach完成后调用$conn -> query($sql)，这意味着在$sql中只存储了foreach循环中的最后一个值。

试试这个：

// Create connection
$conn = new mysqli( $servername, $username, $password, $dbname );
// Check connection
if ( $conn->connect_error) {
    die("Connection failed: ".$conn -> connect_error);
}

foreach ( $_POST['nome'] as $nome ) {

    $nome = $_POST['nome'];
    $cognome = $_POST['cognome'];
    $rimanenti_ins = $_POST['rimanenti'];


//$sql = "UPDATE rimanenti_mese SET Rimanenti='$rimanenti_ins' WHERE Nome='$nome' AND Cognome = '$cognome'";

    $sql = "INSERT INTO rimanenti_mese (Nome, Cognome, Mese, Anno, Benzina, Pezzi, Rimanenti)
VALUES('$nome', '$cognome', '09', '2015', 'xxxx', 'fdfdfd', '$rimanenti_ins')
";

    //put your ->query into the foreach loop
    if ($conn -> query($sql) === TRUE) {
        echo "New record created successfully";
    } else {
        echo "Error: ".$sql.
        "<br>".$conn -> error;
    }
}

$conn -> close();

在PHP中调用post后多次插入

2 个答案: