jquery传递两个数组并插入一个表

Question

我在两个数组中存储了相同形式的标签和数值，使用AJAX将这些数组传递给php。现在我想将每个数组存储在同一个表mysql的两个不同字段中。

＆＃13;

＆＃13;

var arr = [];
var arr1 = [];

$('.cat').each(function() {         
   arr.push($(this).text());
});

$('.cat_value').each(function() {         
   arr1.push($(this).val());
});

$.ajax({
    url:'rs_insert.php',
    data:{categories: arr, cat_values:arr1},
    type:'POST',
    success:function() {
        alert("data has been sent");
		document.getElementById('exampleModal1').style.display = "none"; 
    }      
}); 

＆＃13;

＆＃13;

＆＃13;

php文件

＆＃13;

＆＃13;

$con=mysqli_connect("localhost","root","","test");

// Check connection
if (mysqli_connect_errno()) {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$values = array();
foreach($_POST['categories'] as $key => $name) { 
  $values[] = $name;
}

$sql1 = "INSERT INTO rs (Category) VALUES ";
foreach($values as $cat) {
  $sql1 .= "('" . $cat . "'),";
}

$sql1_trimmed = rtrim($sql1,',');

////

$values1 = array();
foreach($_POST['cat_values'] as $key => $value) { 
  $values1[] = $value;
}

$sql2 = "INSERT INTO rs (Value) VALUES ";
foreach($values1 as $val) {
  $sql2 .= "('" . $val . "'),";
}

$sql2_trimmed = rtrim($sql2,',');
if (!mysqli_query($con,$sql1_trimmed)) {
  die('Error: ' . mysqli_error($con));
}
 if (!mysqli_query($con,$sql2_trimmed)) {
   die('Error: ' . mysqli_error($con));
 }

＆＃13;

＆＃13;

＆＃13;

问题是，它没有插入值并行。首先，它在类别字段中插入值，其中值为0。带有空白类别名称的填充值字段。

Answer 1

你可以试试这个：

$con=mysqli_connect("localhost","root","","test");

// Check connection
if (mysqli_connect_errno()) {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$values = array();
foreach($_POST['categories'] as $key => $name) { 
  $values[] = $name;
}

////

$values1 = array();
foreach($_POST['cat_values'] as $key => $value) { 
  $values1[] = $value;
}

$sql1 = "INSERT INTO rs (Category,Value) VALUES ";
for($i=0;$i<count($values);$i++) {
  $sql1 .= "('" . $values[$i] . "','". $values1[$i] ."'),";
}

$sql1_trimmed = rtrim($sql1,',');

if (!mysqli_query($con,$sql1_trimmed)) {
  die('Error: ' . mysqli_error($con));
}