我想将数组值插入到具有相同联系人ID的数据库中,
我想要这样
contactid languageid
124 1
124 2
这是我的languageid数组值:Array([0] => 1 [1] => 2 [2] =>)和我的contactid = 124
任何人都可以告诉我这是怎么回事,
$queryinsert="INSERT INTO contactlanguage (contactid,languageid) VALUES ('$languageId','$contactId')";
mysql_query($queryinsert);
print $queryinsert;
答案 0 :(得分:2)
您只有列不匹配(您的列已切换)。考虑这个例子:
// provided, you have already connected to mysql
$languageid = array(1, 2);
$contactid = 124;
foreach($languageid as $id) {
$statement = "INSERT INTO contactlanguage (contactid, languageid) VALUES ('$contactid', '$id')"
mysql_query($statement);
}
答案 1 :(得分:0)
$queryinsert="INSERT INTO contactlanguage (contactid,languageid) VALUES ('$languageId','$contactId')";
mysql_query($queryinsert);
print $queryinsert;
你的迭代值是错误的。您传递给查询的值是错误的格式,
检查你的第一个字段是contactid,第二个字段是languageid,你的第一个值是languageid,第二个是contactid,
因此,在这种情况下,您的languageid进入contactid字段,contactid进入languageid字段。所以请按照以下说明,
$languageid = array(xx, xx); // Which is your array values
$contactid = xxx; // which is your conatec value.
foreach($languageid as $key=>$val) {
$statement = "INSERT INTO contactlanguage (contactid, languageid) VALUES ('".$contactid."', '".$val."')"
mysql_query($statement);
}
答案 2 :(得分:0)
试试这个:
$languageid = array(1, 2);
$contactid = 124;
foreach($languageid as $key=>$id) {
$statement = "INSERT INTO contactlanguage (contactid, languageid) VALUES ('$contactid', '$id')"
mysql_query($statement);
}
答案 3 :(得分:0)
首先请记住,mysql_ *函数已被折旧且您为obsolete to sql injection,因为您直接将用户输入传递给查询。是时候切换到mysqli_*或pdo。
迭代你的languageid数组并插入每个这样的数组,你需要在查询中更改为列名。
$arrlanguageid = array ( 0 => 1 , 1 => 2);
$contactid = 124;
foreach($arrlanguageid as $key=>$val){
$queryinsert="INSERT INTO contactlanguage (contactid,languageid) VALUES ('$contactid','$val')";
mysql_query($queryinsert);
}
答案 4 :(得分:0)
你可以这样做
<?php
$con=mysqli_connect("your_db_ip","your_username","your_password","your_db_name");
$array = array(1,2,3);
$contactId = 124;
foreach($array as $value){
$languageId = $value;
$queryinsert="INSERT INTO contactlanguage (contactid,languageid) VALUES ('$contactId','$languageId')";
mysqli_query($con,$queryinsert);
}
?>