我想使用DB在PHP中对我的字符串进行一些替换类型的操作。例如:
$inputString = "My name is [[name-abc]] and age [[age-25]]";
我有如下DB表:
id input output
1 name-abc LINK_TO_ABC_PROFILE
2 name-def LINK_TO_DEF_PROFILE
3 age-18 LINK_TO_AGE_18
4 age-25 LINK_TO_AGE_25
我需要输出:
$outputString = "My name is LINK_TO_ABC_PROFILE and age LINK_TO_AGE_25";
我用preg_replace尝试了各种方法,但没有得到结果。假设DB在我的数组中,有人能为我编写如下函数吗:
array('name-abc' => LINK_TO_ABC_PROFILE, 'name-def' => LINK_TO_DEF_PROFILE .... 'age-25' => LINK_TO_AGE_25)
提前感谢!
$inputString = "My name is [[name-abc]] and age [[age-25]]";
$replace = array('name-abc' => LINK_TO_ABC_PROFILE, 'name-def' => LINK_TO_DEF_PROFILE , 'age-25' => LINK_TO_AGE_25);
$keys = array();
foreach ($replace as $k => $v) {
$keys[] = '[[' . $k . ']]';
}
$out = str_replace ( $keys , array_values($replace) , $inputString );
var_dump(($out));
输出:
string(53) "My name is LINK_TO_ABC_PROFILE and age LINK_TO_AGE_25"
<?php
$details = array('name-abc' => LINK_TO_ABC_PROFILE, 'name-def' => LINK_TO_DEF_PROFILE .... 'age-25' => LINK_TO_AGE_25);
$outputString = "My name is ".$details['name-abc']." and age ".$details['age-25'];
?>
如果你想使用preg_replace,你必须构建两个一维数组,而不是关联数组,试着如下,
$input_arr = array();
$output_arr = array();
$query = "SELECT input,output FROM replacestrtbl";
$stmt = $mysqli->prepare($query);
if($stmt){
$stmt->execute();
$stmt->bind_result($input, $output);
$i=0;
while($res = $stmt->fetch()){
$input_arr[$i] = '/'['['.$input.'']']/';
$output_arr[$i] = $output;
$i++;
}
$stmt->close();
}
$inputString = "My name is [[name-abc]] and age [[age-25]]";
$output_string=preg_replace($input_arr,$output_arr,$inputString);
echo $output_string;