Wil*_*aan 5 php mysql pdo mysql-error-1064
我正在使用mysql,我有一个表,他的名字是"概念关系",我想插入其中
for ($i = 0; $i < count($sources); $i++) {
$sourceID = $this->getConcpeptID($sources[$i]);
$desID = $this->getConcpeptID($distinations[$i]);
$query2 = "INSERT INTO concept-relation (relationID, firstConceptID, secondConceptID) VALUES (:rID, :sID, :dID)";
$sth = $this->db->prepare($query2);
$sth->execute(array(
':rID' => $relationID,
':sID' => $sourceID,
'dID' => $desID
));
}
Run Code Online (Sandbox Code Playgroud)
我收到此语法错误消息
INSERT INTO concept - relation(
relationID,
firstConceptID,
secondConceptID
)
VALUES ( 3, 3, 3 )
Run Code Online (Sandbox Code Playgroud)
我试图直接从mysql插入并得到错误似乎相同的错误
for ($i = 0; $i < count($sources); $i++) {
$sourceID = $this->getConcpeptID($sources[$i]);
$desID = $this->getConcpeptID($distinations[$i]);
$query2 = "INSERT INTO concept-relation (relationID, firstConceptID, secondConceptID) VALUES (:rID, :sID, :dID)";
$sth = $this->db->prepare($query2);
$sth->execute(array(
':rID' => $relationID,
':sID' => $sourceID,
'dID' => $desID
));
}
Run Code Online (Sandbox Code Playgroud)
问题是因为表的名称在ti中有破折号,看看mysql如何理解查询
INSERT INTO concept - relation(
relationID,
firstConceptID,
secondConceptID
)
VALUES ( 3, 3, 3 )
Run Code Online (Sandbox Code Playgroud)
它理解概念,只做" - 关系",
请帮忙
但不改变我的表名:)
Shi*_*dim 13
用反引号括起标识符会使保留字/字符成为mysql中的有效标识符.
所以你应该使用
`concept-relation`
Run Code Online (Sandbox Code Playgroud)
反引号(`)是此键盘左上角的键.
