Question

我正在使用jscolor.js函数来显示颜色选择器。用户选择一种颜色，然后将其输入mysql供以后使用。问题是我可以回显/提醒所选颜色，但它没有传递给php。如果我硬编码它工作正常。任何人都对我遗失的内容有所建议。

<?php

include ('config.php');

$get_coords = mysqli_query($link, "SELECT * FROM color WHERE color_id = '1'");
while($row = mysqli_fetch_array($get_coords)) {
$color_id = $row['color_id'];
$color = $row['color'];

echo "<div style=color:#$color;>Select a color:</div> <input id='myField1'>";
}

?>

<script type="text/javascript">

$(document).ready(function() {

var myPicker = new jscolor.color(document.getElementById('myField1'), {});
myPicker.fromString('<?php echo $color; ?>');

$("#myField1").change(function() {
alert(myPicker);
$.post('updatecolor.php', { 'color': myPicker });
});
});

</script>

updatecolor.php

<?php

$link = mysqli_connect($db_host, $db_usr, $db_pass) or die("MySQL Error: " . mysqli_error());

mysqli_select_db($link, $db_name) or die("MySQL Error: " . mysqli_error());

$color=$_POST['color'];

$sql = "UPDATE color SET color = '$color' WHERE color_id = '1'";

mysqli_query($link, $sql) or die("Error updating Coords :".mysqli_error());

echo "success";

?>

Answer 1

现在全部排序。

我已将PHP更改为：

echo "<div id='mypick' style=color:#$color;><h1>Select a color:</h1></div>";

并将JQuery更改为：

var myPicker = $("#myField1").val();
$.post('updatecolor.php', { color:myPicker });
document.getElementById('mypick').style.color = myPicker;

将jscolor选择变量从jquery传递给php无法正常工作

1 个答案: