因某些原因无法传递变量？ PHP

Question

Some reason when i call $_POST['xalue'] it will only return the first index of $xaxis.

$xaxis = '["-100","-100","240","1056","3560","-650","5980.5","1020"]'

im just trying to pass the information so when the button is pressed to add the dashboard i can pass the data along. Thank you

$query = mysqli_query($db, "SELECT ID From users WHERE Username='$user'");
$fetch = mysqli_fetch_assoc($query);
$data = $fetch{"ID"};

$sql = "SELECT dashboardname FROM dashboards WHERE dashboarduserID=$data";
$result = $db->query($sql);
$addarray = array();
if ($result->num_rows > 0) {
    global $xaxis;
    echo $xaxis;
    while ($row = $result->fetch_assoc()) {
        array_push($addarray, $row['dashboardname']);
    }
    echo '<form method="post">';?>
    <input type="text" name="xvalue" value="<?php echo $xaxis ?>">
    <?php
    echo 'Add current graph to :';
    echo '<select name="addgraph">';
    foreach ($addarray as $item) {
        echo '<option value="' . $item . '">' . $item . '</option>';
    }
    echo '</select>';
    echo '<input type="submit" name="addtodash" value="Add">';
    echo '</form>';
} else {
    echo "0 results";
    if (isset($_REQUEST["addtodash"])) {
    $xaxis = $_POST['xvalue'];
    $query = mysqli_query($db, "SELECT ID From users WHERE Username='$user'");
    $fetch = mysqli_fetch_assoc($query);
    $data = $fetch{"ID"};
    $nameofdash = $_POST['addgraph'];
    mysqli_query($db, "INSERT INTO dashboardsdata (userid,dashboardname,xdata,ydata) VALUES ($data,\"$nameofdash\",\"$xaxis\",\"$yaxis\");");
    echo "done";
}

Answer 1

The double quotes in $xvalue will match the double quotes that are used to delimit the value="..." attribute. You need to encode them:

<input type="text" name="xvalue" value="<?php echo htmlspecialchars($xaxis); ?>">