SQL更新
问题描述:
比方说,我已经和数组这样SQL更新
$array= Array('id'=>'3', 'name'=>'NAME', 'age'=>'12');
从这个数组键是表和值的列的名称是,我需要更新的列值。 我想根据键和值更新表。 我使用ADODB 请帮我
答
你可以使用这样的实现是:
foreach($values as $value) {
if(!key_exists($value, $item)) {
return false;
}
$table->{$value} = $items[$value];
}
答
试试这个:
$sql = "UPDATE table SET ";
foreach($array as $key=>$value) {
$sql .= $key . " = " . $value . ", ";
}
$sql = trim($sql, ' '); // first trim last space
$sql = trim($sql, ','); // then trim trailing and prefixing commas
,当然还有WHERE
条款:
$sql .= " WHERE condition = value";
你将获得这种严格G:
UPDATE table SET id = 3, name = NAME, age = 12 WHERE condition = value
LE:您可能需要添加引号为字符串,所以我有我的代码更改为类似这样:
$sql = "UPDATE table SET ";
foreach($array as $key=>$value) {
if(is_numeric($value))
$sql .= $key . " = " . $value . ", ";
else
$sql .= $key . " = " . "'" . $value . "'" . ", ";
}
$sql = trim($sql, ' '); // first trim last space
$sql = trim($sql, ','); // then trim trailing and prefixing commas
$sql .= " WHERE condition = value";
这会产生这样的:
UPDATE table SET id = 3, name = 'NAME', age = 12 WHERE condition = value
LE 2:如果您想要条件中的id列,代码将变为:
$sql = "UPDATE table SET ";
foreach($array as $key=>$value) {
if($key == 'id'){
$sql_condition = " WHERE " . $key . " = " . $value;
continue;
}
if(is_numeric($value))
$sql .= $key . " = " . $value . ", ";
else
$sql .= $key . " = " . "'" . $value . "'" . ", ";
}
$sql = trim($sql, ' '); // first trim last space
$sql = trim($sql, ','); // then trim trailing and prefixing commas
$sql .= $sql_condition;
这将产生这样的结果:
UPDATE table SET name = 'NAME', age = 12 WHERE id = 3
希望这有助于! :d
答
假设关键指标始终ID和ADODB可以使用指定的占位符,你可以这样做:
$array = Array('id'=>'3', 'name'=>'NAME', 'age'=>'12');
$set = array();
$data = array();
while(list($key,$value)=each($array)) {
$data[':'.$key] = $value;
if($key!='id') {
$set[] = $key . ' = :' . $key;
// if no placeholders use $set[] = $key . " = '" . database_escape_function($value) . "'";
}
}
$sql = "UPDATE table SET ".implode($set, ',')." WHERE id=:id";
//$data is now Array(':id'=>'3', ':name'=>'NAME', ':age'=>'12');
//$sql is now "UPDATE table SET name=:name, age=:age WHERE id=:id";
$stmt = $DB->Prepare($sql);
$stmt = $DB->Execute($stmt, $data);
检查http://stackoverflow.com/questions/23945476/php-sql-update - 阵列和http://stackoverflow.com/questions/7884284/simple-update-mysql-table-from-php-array – Kavvson 2014-11-04 13:03:12
以下答案任何工作的人吗? – 2014-11-04 13:35:01